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Introduction to Management Science A Modeling and Case Studies Approach with Spreadsheets The McGraw-Hill/Irwin Series in Operations and Decision Sciences SUPPLY CHAIN MANAGEMENT BUSINESS FORECASTING BUSINESS MATH Benton Purchasing and Supply Chain Management Second Edition Keating and Wilson Business Forecasting Seventh Edition Slater and Wittry Practical Business Math Procedures Twelfth Edition Burt, Petcavage, and Pinkerton Supply Management Eighth Edition Bowersox, Closs, Cooper, and Bowersox Supply Chain Logistics Management Fourth Edition Johnson and Flynn Purchasing and Supply Management Fifteenth Edition Simchi-Levi, Kaminsky, and Simchi-Levi Designing and Managing the Supply Chain: Concepts, Strategies, Case Studies Third Edition PROJECT MANAGEMENT Brown and Hyer Managing Projects: A Team-Based Approach First Edition Larson and Gray Project Management: The Managerial Process Seventh Edition SERVICE OPERATIONS MANAGEMENT Bordoloi, Fitzsimmons, and Fitzsimmons Service Management: Operations, Strategy, Information Technology Ninth Edition MANAGEMENT SCIENCE Hillier and Hillier Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets Sixth Edition Stevenson and Ozgur Introduction to Management Science with Spreadsheets First Edition MANUFACTURING CONTROL SYSTEMS Jacobs, Berry, Whybark, and Vollmann Manufacturing Planning & Control for Supply Chain Management Sixth Edition BUSINESS RESEARCH METHODS Cooper and Schindler Business Research Methods Twelfth Edition LINEAR STATISTICS AND REGRESSION Kutner, Nachtsheim, and Neter Applied Linear Regression Models Fourth Edition BUSINESS SYSTEMS DYNAMICS Sterman Business Dynamics: Systems Thinking and Modeling for a Complex World First Edition OPERATIONS MANAGEMENT Cachon and Terwiesch Operations Management First Edition Cachon and Terwiesch Matching Supply with Demand: An Introduction to Operations Management Third Edition Finch Interactive Models for Operations and Supply Chain Management First Edition Jacobs and Chase Operations and Supply Chain Management Fifteenth Edition Jacobs and Chase Operations and Supply Chain Management: The Core Fourth Edition Jacobs and Whybark Why ERP? A Primer on SAP Implementation First Edition Schroeder and Goldstein Operations Management in the Supply Chain: Decisions and Cases Seventh Edition Stevenson Operations Management Thirteenth Edition Swink, Melnyk, Hartley, and Cooper Managing Operations Across the Supply Chain Third Edition Slater and Wittry Math for Business and Finance: An Algebraic Approach Second Edition BUSINESS STATISTICS Bowerman, O’Connell, and Murphree Business Statistics in Practice Eighth Edition Bowerman, O’Connell, Murphree, and Orris Essentials of Business Statistics Fifth Edition Doane and Seward Applied Statistics in Business and Economics Sixth Edition Doane and Seward Essential Statistics in Business and Economics Second Edition Lind, Marchal, and Wathen Basic Statistics for Business and Economics Ninth Edition Lind, Marchal, and Wathen Statistical Techniques in Business and Economics Seventeenth Edition Jaggia and Kelly Business Statistics: Communicating with Numbers Third Edition Jaggia and Kelly Essentials of Business Statistics: Communicating with Numbers First Edition McGuckian Connect Master: Business Statistics Introduction to Management Science Introduction to Management Science A Modeling and Case Studies Approach with Spreadsheets Sixth Edition Frederick S. Hillier Stanford University Mark S. Hillier University of Washington Cases developed by Karl Schmedders University of Zurich Molly Stephens Quinn, Emanuel, Urquhart & Sullivan, LLP Final PDF to printer INTRODUCTION TO MANAGEMENT SCIENCE Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2019 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 QVS 21 20 19 18 ISBN MHID 978-1-260-09185-4 1-260-09185-6 Cover Image: ©polygraphus/Getty Images All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites. mheducation.com/highered hiL91856_fm_ISE.indd iv 12/11/17 02:27 PM To the memory of Christine Phillips Hillier a beloved wife and daughter-in-law Gerald J. Lieberman an admired mentor and one of the true giants of our field About the Authors Frederick S. Hillier is professor emeritus of operations research at Stanford University. Dr. Hillier is especially known for his classic, award-winning text, Introduction to Operations Research, co-authored with the late Gerald J. Lieberman, which has been translated into well over a dozen languages and is currently in its 10th edition. The 6th edition won honorable mention for the 1995 Lanchester Prize (best English-language publication of any kind in the field), and Dr. Hillier also was awarded the 2004 INFORMS Expository Writing Award for the 8th edition. His other books include The Evaluation of Risky Interrelated Investments, Queueing Tables and Graphs, Introduction to Stochastic Models in Operations Research, and Introduction to Mathematical Programming. He received his BS in industrial engineering and doctorate specializing in operations research and management science from Stanford University. The winner of many awards in high school and college for writing, mathematics, debate, and music, he ranked first in his undergraduate engineering class and was awarded three national f­ ellowships (National Science Foundation, Tau Beta Pi, and Danforth) for graduate study. After receiving his PhD degree, he joined the faculty of Stanford University, where he earned tenure at the age of 28 and the rank of full professor at 32. Dr. Hillier’s research has extended into a variety of areas, including integer programming, queueing theory and its application, statistical quality control, and production and operations management. He also has won a major prize for research in capital budgeting. Twice elected a national officer of professional societies, he has served in many important professional and editorial capacities. For example, he served The Institute of Management Sciences as vice president for meetings, chairman of the publications committee, associate editor of Management Science, and co-general chairman of an international conference in Japan. He also is a Fellow of the Institute for Operations Research and the Management Sciences (INFORMS). He served for 20 years (until 2013) as the founding series editor for a prominent book series, the International Series in Operations Research and ­Management ­Science, for Springer Science + Business Media. He has had visiting appointments at Cornell University, the Graduate School of Industrial Administration of Carnegie-Mellon University, the Technical University of Denmark, the University of Canterbury (New Zealand), and the Judge Institute of Management Studies at the University of Cambridge (England). vi Mark S. Hillier, son of Fred Hillier, is associate professor of quantitative methods at the Michael G. Foster School of Business at the University of Washington. Dr. Hillier received his BS in engineering (plus a concentration in computer science) from Swarthmore College. He then received his MS with distinction in operations research and PhD in industrial engineering and engineering management from Stanford University. As an undergraduate, he won the McCabe Award for ranking first in his engineering class, won election to Phi Beta Kappa based on his work in mathematics, set school records on the men’s swim team, and was awarded two national fellowships (National Science Foundation and Tau Beta Pi) for graduate study. During that time, he also developed a comprehensive software tutorial package, OR Courseware, for the Hillier–Lieberman textbook, Introduction to Operations Research. As a graduate student, he taught a PhD-level seminar in operations management at Stanford and won a national prize for work based on his PhD dissertation. At the University of Washington, he currently teaches courses in management science and spreadsheet modeling. He has won over twenty MBA teaching awards for the core course in management science and his elective course in spreadsheet modeling, as well as a universitywide teaching award for his work in teaching undergraduate classes in operations management. He was chosen by MBA students in both 2007 and 2013 as the winner of the prestigious PACCAR award for Teacher of the Year (reputed to provide the largest monetary award for MBA teaching in the nation). He won the Ron Crocket Award for Innovation in Education in 2014. He also has been awarded an appointment to the Evert McCabe Endowed Faculty Fellowship. His research interests include issues in component commonality, inventory, manufacturing, and the design of production systems. A paper by Dr. Hillier on component commonality won an award for best paper of 2000–2001 in IIE Transactions. He also has been serving as principal investigator on a grant from the Bill and Melinda Gates Foundation to lead student research projects that apply spreadsheet modeling to various issues in global health being studied by the foundation. About the Case Writers Karl Schmedders is professor of quantitative business administration at the University of Zurich in Switzerland and a visiting professor of executive education at the Kellogg School of Management of Northwestern University. His research interests include management science, business analytics, and computational economics and finance. He received his PhD in operations research from Stanford University, where he taught both undergraduate and graduate classes in management science, including a case studies course. He received several teaching awards at Stanford, including the universitywide Walter J. Gores Teaching Award. After a post-doc at the Hoover Institution, a think tank on the Stanford campus, he became assistant professor of managerial economics and decision sciences at the Kellogg School. He was promoted to associate professor in 2001 and received tenure in 2005. In 2008 he joined the University of Zurich, where he currently teaches courses in management science, business analytics, and computational economics and finance. He has published research articles in international academic journals such as Management Science, Operations Research, Econometrica, The Review of Economic Studies, and The Journal of Finance, among others. He is the area editor of the field “Computational Economics” for the INFORMS journal Operations Research. At Kellogg he received several teaching awards, including the L. G. Lavengood Professor of the Year Award. Most recently he won the best professor award of the Kellogg School’s European EMBA program (2011, 2012, 2013, 2014, 2015) and its EMBA program in Hong Kong (2017). Molly Stephens is a partner in the Los Angeles office of Quinn, Emanuel, Urquhart & ­Sullivan, LLP. She graduated from Stanford with a BS in industrial engineering and an MS in operations research. Ms. Stephens taught public speaking in Stanford’s School of E ­ ngineering and served as a teaching assistant for a case studies course in management science. As a teaching assistant, she analyzed management science problems encountered in the real world and transformed these into classroom case studies. Her research was rewarded when she won an undergraduate research grant from Stanford to continue her work and was invited to speak at INFORMS to present her conclusions regarding successful classroom case studies. ­Following graduation, Ms. Stephens worked at Andersen Consulting as a systems integrator, experiencing real cases from the inside, before resuming her graduate studies to earn a JD degree with honors from the University of Texas School of Law at Austin. She is a partner in the largest law firm in the United States devoted solely to business litigation, where her practice focuses on complex financial and securities litigation. She also is ranked as a leading securities litigator by Chambers USA (2013 and 2014), which acknowledged “praise for her powerful and impressive securities litigation practice” and noted that she is “phenomenally bright, a critical thinker and great listener.” vii Preface We have long been concerned that traditional management science textbooks have not taken the best approach in introducing business students to this exciting field. Our goal when initially developing this book during the late 1990s was to break out of the old mold and present new and innovative ways of teaching management science more effectively. We have been gratified by the favorable response to our efforts. Many reviewers and other users of the first five editions of the book have expressed appreciation for its various distinctive features, as well as for its clear presentation at just the right level for their business students. Our goal for this sixth edition has been to build on the strengths of the first five editions. Co-author Mark Hillier has won over twenty schoolwide teaching awards for his spreadsheet modeling and management science courses at the University of Washington while using the first five editions, and this experience has led to many improvements in the current edition. We also incorporated many user comments and suggestions. Throughout this process, we took painstaking care to enhance the quality of the preceding edition while maintaining the distinctive orientation of the book. This distinctive orientation is one that closely follows the recommendations in the 1996 report of the operating subcommittee of the INFORMS Business School Education Task Force, including the following extract. There is clear evidence that there must be a major change in the character of the (introductory management science) course in this environment. There is little patience with courses centered on algorithms. Instead, the demand is for courses that focus on business situations, include prominent non-mathematical issues, use spreadsheets, and involve model formulation and assessment more than model structuring. Such a course requires new teaching materials. This book is designed to provide the teaching materials for such a course. In line with the recommendations of this task force, which continue to be widely accepted today, we believe that a modern introductory management science textbook should have three key elements. As summarized in the subtitle of this book, these elements are a modeling and case studies approach with spreadsheets. SPREADSHEETS The modern approach to the teaching of management science clearly is to use spreadsheets as a primary medium of instruction. Both business students and managers now live with spreadsheets, so they provide a comfortable and enjoyable learning environment. Modern spreadsheet software, including Microsoft Excel used in this book, now can be used to do real management science. For student-scale models (which include many practical real-world models), spreadsheets are a much better way of implementing management science models than traditional algebraic solvers. This means that the algebraic curtain that continues to be prevalent in traditional management science courses and textbooks now can be lifted. However, with the current enthusiasm for spreadsheets, there is a danger of going overboard. Spreadsheets are not the only useful tool for performing management science analyses. Occasional modest use of algebraic and graphical analyses still have their place and we would be doing a disservice to the students by not developing their skills in these areas when appropriate. Furthermore, the book should not be mainly a spreadsheet cookbook that focuses largely on spreadsheet mechanics. Spreadsheets are a means to an end, not an end in themselves. A MODELING APPROACH This brings us to the second key feature of the book, a modeling approach. Model formulation lies at the heart of management science methodology. Therefore, we heavily emphasize the art of model formulation, the role of a model, and the analysis of model results. We primarily (but not exclusively) use a spreadsheet format rather than algebra for formulating and presenting a model. viii Preface ix Some instructors have many years of experience in teaching modeling in terms of formulating algebraic models (or what the INFORMS Task Force called “model structuring”). Some of these instructors feel that students should do their modeling in this way and then transfer the model to a spreadsheet simply to use the Excel Solver to solve the model. We disagree with this approach. Our experience (and the experience reported by many others) is that most business students find it more natural and comfortable to do their modeling directly in a spreadsheet. Furthermore, by using the best spreadsheet modeling techniques (as presented in this edition), formulating a spreadsheet model tends to be considerably more efficient and transparent than formulating an algebraic model. Another benefit is that the spreadsheet model includes all the relationships that can be expressed in an algebraic form and we often will summarize the model in this format as well. Another break from past tradition in this book (and several contemporary textbooks) is to virtually ignore the algorithms that are used to solve the models. We feel that there is no good reason why typical business students should learn the details of algorithms executed by computers. Within the time constraints of a one-term management science course, there are far more important lessons to be learned. Therefore, the focus in this book is on what we believe are these far more important lessons. High on this list is the art of modeling managerial problems on a spreadsheet. We believe that training business students in spreadsheet modeling will provide them with two key benefits when they later become managers. First, this will give them a powerful tool for analyzing small managerial problems without requiring outside help. Second, this will enable them to recognize when a management science team could be very helpful for analyzing more complicated managerial problems. Formulating a spreadsheet model of a real problem typically involves much more than designing the spreadsheet and entering the data. Therefore, we work through the process step by step: understand the unstructured problem, verbally develop some structure for the problem, gather the data, express the relationships in quantitative terms, and then lay out the spreadsheet model. The structured approach highlights the typical components of the model (the data, the decisions to be made, the constraints, and the measure of performance) and the different types of spreadsheet cells used for each. Consequently, the emphasis is on the modeling rather than spreadsheet mechanics. A CASE STUDIES APPROACH However, all this still would be quite sterile if we simply presented a long series of brief examples with their spreadsheet formulations. This leads to the third key feature of this book—a case studies approach. In addition to examples, nearly every chapter includes one or two case studies patterned after actual applications to convey the whole process of applying management science. In a few instances, the entire chapter revolves around a case study. By drawing the student into the story, we have designed each case study to bring that chapter’s technique to life in a context that vividly illustrates the relevance of the technique for aiding managerial decision making. This storytelling, case-centered approach should make the material more enjoyable and stimulating while also conveying the practical considerations that are key factors in applying management science. We have been pleased to have several reviewers of the first five editions express particular appreciation for our case study approach. Even though this storytelling approach has received little use in other management science textbooks, we feel that it is a real key to preparing students for the practical application of management science in all its aspects. Some of the reviewers have highlighted the effectiveness of the dialogue/scenario enactment approach used in some of the case studies. Although unconventional, this approach provides a way of demonstrating the process of managerial decision making with the help of management science. It also enables previewing some key concepts in the language of management. Every chapter also contains full-fledged cases following the problems at the end of the chapter. These cases usually continue to employ a stimulating storytelling approach, so they can be assigned as interesting and challenging projects. Most of these cases were developed x Preface jointly by two talented case writers, Karl Schmedders (a faculty member at the University of Zurich in Switzerland) and Molly Stephens (formerly a management science consultant with Andersen Consulting). The authors also have added some cases, including several shorter ones. In addition, the University of Western Ontario Ivey School of Business (the secondlargest producer of teaching cases in the world) has specially selected cases from their case collection that match the chapters in this textbook. These cases are available on the Ivey website, cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. This website address is provided at the end of each chapter as well. We are, of course, not the first to incorporate any of these key features into a management science textbook. However, we believe that the book currently is unique in the way that it fully incorporates all three key features together. OTHER SPECIAL FEATURES We also should mention some additional special features of the book that are continued from the fifth edition. • Diverse examples, problems, and cases convey the pervasive relevance of management science. • A strong managerial perspective. • Learning objectives at the beginning of each chapter. • Numerous margin notes that clarify and highlight key points. • Excel tips interspersed among the margin notes. • Review questions at the end of each section. • A glossary at the end of each chapter. • Partial answers to selected problems in the back of the book. • Extensive supplementary text materials (nine supplements to book chapters and seven additional chapters) are available in the Instructor Resources of McGraw-Hill’s online homework management platform, Connect, (described later) and on the website, www.mhhe .com/Hillier6e. A NEW SOFTWARE PACKAGE This edition continues to integrate Excel and its Solver (a product originally created by Frontline Systems) throughout the book. However, we are excited to also add to this edition an impressive more recent product of Frontline Systems called Analytic Solver ® for ­Education (hereafter referred to as Analytic Solver). Analytic Solver can be used with Microsoft Excel for Windows (as an add-in), or ‘in the cloud” at AnalyticSolver.com using any device (PC, Mac, tablet) with a web browser. It offers comprehensive features for prescriptive analytics (optimization, simulation, decision analysis) and predictive analytics (forecasting, data mining, text mining). Its optimization features are upward compatible from the standard Solver in Excel. Analytic Solver includes: • A more interactive user interface, with the model parameters always visible alongside the main spreadsheet, rather than only in the Solver dialog box. • Parameter analysis reports that provide an easy way to see the effect of varying data in a model in a systematic way. • A model analysis tool that reveals the characteristics of a model (e.g., whether it is linear or nonlinear, smooth or nonsmooth). • Tools to build and solve decision trees within a spreadsheet. • A full range of time series forecasting and data mining models. • The ability to build and run sophisticated Monte Carlo simulation models. • An interactive simulation mode that allows simulation results to be shown instantly whenever a change is made to a simulation model. • The Solver in Analytic Solver can be used in combination with computer simulation to perform simulation optimization. Preface xi • If interested in having students get individual licenses for class use, instructors should send an email to support@solver.com to get their course code and receive student pricing and access information as well as their own access information. Note that this software is no longer free with the purchase of this text, but low-cost student licenses are available. A CONTINUING FOCUS ON EXCEL AND ITS SOLVER As with all the preceding editions, this edition continues to focus on spreadsheet modeling in an Excel format. Although it lacks some of the functionalities of Analytic Solver, the Excel Solver continues to provide a completely satisfactory way of solving most of the spreadsheet models encountered in this book. This edition continues to feature this use of the Excel Solver whenever either it or the Analytic Solver could be used. Many instructors prefer this focus because it avoids introducing other complications that might confuse their students. We agree. However, the key advantage of introducing Analytic Solver in this edition is that it provides an all-in-one complement to the Excel Solver. There are some important topics in the book (including decision analysis and computer simulation) where the Excel Solver lacks the functionalities needed to deal fully with these kinds of problems. Multiple Excel add-ins— Solver Table, TreePlan, SensIt, RiskSim, Crystal Ball, and OptQuest (a module of Crystal Ball)—were introduced in previous editions to provide the needed functionalities. Analytic Solver alone now replaces all of these add-ins. To further enhance a continuing focus on Excel and its Solver, www.mhhe.com/Hillier6e includes all the Excel files that provide the live spreadsheets for all the various examples and case studies throughout the book. In addition to further investigating the examples and case studies, these spreadsheets can be used by either the student or instructor as templates to formulate and solve similar problems. This website also includes dozens of Excel templates for solving various models in the book as well as a Queueing Simulator for performing computer simulations of queueing systems (used in Chapter 12). NEW FEATURES IN THIS EDITION We have made some important enhancements to the current edition. • A New Section Describes the Relationship Between Analytics and Management Science. Recent years have seen an exciting analytics revolution as the business world has come to recognize the key role that analytics can play in managerial decision making. A new ­Section 1.3 fully describes the close relationship between analytics and management science. • A New Section on the Role of Robust Optimization in What-If Analysis. The goal of robust optimization is to find a solution for a model that is virtually guaranteed to remain feasible and near optimal for all plausible combinations of the actual values for the parameters of the model. A new Section 5.7 describes this key tool for performing what-if analysis for linear programming models. • A New Section on Chance Constraints. Constraints in a model usually are fully required to be satisfied, but occasionally constraints arise that actually have a little flexibility. Chance constraints provide a way of dealing with such constraints that actually can be violated a little bit without serious complications. A new Section 5.8 describes the role of chance constraints and how to implement them. • A Thorough Updating Throughout the Book. Given that the writing of the first edition occurred approximately 20 years ago, it is inevitable that some of that writing now is somewhat outdated. Even though the descriptions of management science techniques may still be accurate, the numbers and other details describing their application in certain problems, cases, and examples may now seem quite obsolete. Wage standards have changed. Prices have changed. Technologies have changed. Dates have changed. Although we did some updating with each new edition, we made a special effort this time to thoroughly update numbers and other details as needed to reflect conditions in 2017. • Additional Links to Articles that Describe Dramatic Real Applications. The fifth edition included 28 application vignettes that described in a few paragraphs how an actual application of management science had a powerful effect on a company or organization by using xii Preface techniques like those being studied in that portion of the book. The current edition adds five more vignettes based on recent applications (while deleting seven outdated ones) and also updates the information in nine of the other vignettes. We continue the practice of adding a link to the journal articles that fully describe these applications (except that the vignette for Chapter 1 doesn’t require a link), through a special arrangement with the Institute for Operations Research and the Management Sciences (INFORMS®). Thus, the instructor now can motivate his or her lectures by having the students delve into real applications that dramatically demonstrate the relevance of the material being covered in the lectures. The end-ofchapter problems also include an assignment after reading each of these articles. We continue to be excited about this partnership with INFORMS, our field’s preeminent professional society, to provide a link to each of these articles describing spectacular applications of management science. INFORMS is a learned professional society for students, academics, and practitioners in analytics, operations research, and management science. Information about INFORMS journals, meetings, job bank, scholarships, awards, and teaching materials is available at www.informs.org. • A Word-by-Word Review to Further Increase Clarity in Each Chapter. A hallmark of each edition has been a particularly heavy use of certain techniques to maximize the clarity of the material: use cases to bring the material to life, divide sections into smaller subsections, use short paragraphs, use bullet points, set off special conclusions, use i­talics or boldface to highlight key points, add margin notes, never assume too much about ­understanding preceding material, etc. However, we have doubled down with this approach in the current edition by using a word-by-word review of each chapter to further increase clarity while also taking into special account the input provided by reviewers and others. McGRAW-HILL CONNECT(R) LEARN WITHOUT LIMITS! Connect is a teaching and learning platform that is proven to deliver better results for students and instructors. Connect empowers students by continually adapting to deliver precisely what they need, when they need it, and how they need it, so your class time is more engaging and effective. New to the 6th edition, Connect includes multiple-choice questions for each chapter to be used as practice or homework for students, SmartBook(R), instructor resources (including the test bank), and student resources. For access, visit connect.mheducation.com or contact your McGraw-Hill sales representative. SMARTBOOK(R) Proven to help students improve grades and study more efficiently, SmartBook contains the same content within the print book, but actively tailors that content to the needs of the individual. SmartBook’s adaptive technology provides precise, personalized instruction on what the student should do next, guiding the student to master and remember key concepts, targeting gaps in knowledge and offering customized feedback, and driving the student toward comprehension and retention of the subject matter. Available on desktops and tablets, SmartBook puts learning at the student’s fingertips—anywhere, anytime. INSTRUCTOR RESOURCES The Instructor Resource Library within Connect is password-protected and a convenient place for instructors to access course supplements that include nine supplements to chapters in the print book and seven supplementary chapters. Resources for professors include the complete solutions to all problems and cases, PowerPoint slides which include both lecture materials for nearly every chapter and nearly all the figures (including all the spreadsheets) in the book, and an expanded Test Bank. The test bank contains almost 1,000 multiple-choice and true-false questions, all tagged according to learning objective, topic, level of difficulty, Bloom’s taxonomy and AACSB category for filtering and reporting, and in delivered in the following ways: - As a Connect assignment for online testing and automatic grading; can be used for actual exams or assigned as quizzes or practice; - In TestGen, a desktop test generator and editing application for instructors to provide printed tests that can incorporate both McGraw-Hill’s and instructors’ questions; - As Word files, with both question-only and answer files. Final PDF to printer Preface xiii STUDENT RESOURCES As described above, SmartBook provides a powerful tool to students for personalized instruction. For the additional convenience of students, we also are providing the website, www.mhhe.com/ Hillier6e, to provide the full range of resources of interest to students. In addition to providing access to supplementary text material (both supplements to book chapters and additional chapters), this website provides solutions to the “solved problems” (additional examples) that are included at the end of each chapter. For each spreadsheet example in the book, a live spreadsheet that shows the formulation and solution for the example also is provided in the website for easy reference and for use as a template. (At the end of each chapter, the page entitled “Learning Aids for This Chapter” lists the Excel files and other resources that are relevant for that chapter.) Information about accessing the book’s software is provided. In addition, the website includes a tutorial with sample test questions (different from those in the instructor’s test bank) for selftesting quizzes on the various chapters. It also provides access to the INFORMS articles cited in the application vignettes as well as updates about the book, including errata. AN INVITATION We invite your comments, suggestions, and errata. You can contact either one of us at the e-mail addresses given below. While giving these addresses, let us also assure instructors that we will continue our policy of not providing solutions to problems and cases in the book to anyone (including your students) who contacts us. We hope that you enjoy the book. Frederick S. Hillier Stanford University (fhillier@stanford.edu) Mark S. Hillier University of Washington (mhillier@uw.edu) June 2017 Acknowledgments This new edition has benefited greatly from the sage advice of many individuals. To begin, we would like to express our deep appreciation to the following individuals who provided formal reviews of the fifth edition: Michael Cervertti, University of Memphis Jose H. Dula, Virginia Commonwealth University John H. Newman, Coppin State University Danxia Chen, Dallas Baptist University Kenneth Donald Lawrence, New Jersey Institute of Technology David Snowden, Thomas More College Moula Cherikh, Winston-Salem State University Nicoleta Maghear, Hampton University John Wang, Montclair State University Kelwyn A. D’Souza, Hampton University William P. Millhiser, Baruch College, City University of New York Norman Douglas Ward, University of Mount Olive We also are grateful for the valuable input provided by many of our students as well as various other students and instructors who contacted us via e-mail. This book has continued to be a team effort involving far more than the two coauthors. As a third coauthor for the first edition, the late Gerald J. Lieberman provided important initial impetus for this project. We also are indebted to our case writers, Karl Schmedders and Molly Stephens, for their invaluable contributions. Ann Hillier again devoted numerous hours to sitting with a Macintosh, doing word processing and constructing figures and tables. They all were vital members of the team. McGraw-Hill/Irwin’s editorial and production staff provided the other key members of the team, including Noelle Bathurst, Portfolio Manager; Allison McCabe, and Tobi Philips, Product Developers, Harper Christopher, Marketing Manager, and Daryl Horrocks, Program Manager. This book is a much better product because of their guidance and hard work. It has been a real pleasure working with such a thoroughly professional staff. hiL18920_fm_i-xviii.indd xiii 12/20/17 09:44 AM Brief Contents 1 Introduction 1 10 Forecasting 397 2 Linear Programming: Basic Concepts 24 11 Queueing Models 446 3 Linear Programming: Formulation and Applications 64 12 Computer Simulation: Basic Concepts 499 4 The Art of Modeling with Spreadsheets 124 5 What-If Analysis for Linear Programming 151 6 Network Optimization Problems 203 7 Using Binary Integer Programming to Deal with Yes-or-No Decisions 243 8 Nonlinear Programming 279 13 Computer Simulation with Analytic Solver 536 APPENDIXES A Tips for Using Microsoft Excel for Modeling 606 B Partial Answers to Selected Problems 612 INDEX 616 9 Decision Analysis 334 CHAPTERS available at www.mhhe.com/Hillier6e 14 Solution Concepts for Linear Programming 15 Transportation and Assignment Problems 16 PERT/CPM Models for Project Management 17 Goal Programming xiv 18 Inventory Management with Known Demand 19 Inventory Management with Uncertain Demand 20 Computer Simulation with Crystal Ball Contents Chapter One Introduction 1 The Nature of Management Science 2 An Illustration of the Management Science Approach: Break-Even Analysis 6 1.3 The Relationship Between Analytics and Management Science 12 1.4 The Impact of Management Science 14 1.5 Some Special Features of this Book 18 1.6 Summary 19 Glossary 20 Learning Aids for This Chapter 20 Solved Problem 21 Problems 21 Case 1-1 Keeping Time 23 1.1 1.2 Chapter Two Linear Programming: Basic Concepts 24 A Case Study: The Wyndor Glass Co. Product-Mix Problem 25 2.2 Formulating the Wyndor Problem on a Spreadsheet 27 2.3 The Mathematical Model in the Spreadsheet 33 2.4 The Graphical Method for Solving Two-Variable Problems 35 2.5 Using Excel’s Solver to Solve Linear Programming Problems 39 2.6 Analytic Solver 43 2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 47 2.8 Linear Programming from a Broader Perspective 52 2.9 Summary 54 Glossary 54 Learning Aids for This Chapter 55 Solved Problems 55 Problems 55 Case 2-1 Auto Assembly 60 Case 2-2 Cutting Cafeteria Costs 61 Case 2-3 Staffing a Call Center 62 2.1 Supplement to Chapter 2: More about the Graphical Method for Linear Programming (This supplement is available at www.mhhe.com/Hillier6e). Chapter Three Linear Programming: Formulation and Applications 64 3.1 3.2 3.3 3.4 3.5 3.6 A Case Study: The Super Grain Corp. Advertising-Mix Problem 65 Resource-Allocation Problems 71 Cost–Benefit–Trade-Off Problems 81 Mixed Problems 87 Transportation Problems 95 Assignment Problems 99 3.7 Model Formulation from a Broader Perspective 102 3.8 Summary 104 Glossary 104 Learning Aids for This Chapter 104 Solved Problems 105 Problems 106 Case 3-1 Shipping Wood to Market 114 Case 3-2 Capacity Concerns 115 Case 3-3 Fabrics and Fall Fashions 117 Case 3-4 New Frontiers 118 Case 3-5 Assigning Students to Schools 119 Case 3-6 Reclaiming Solid Wastes 120 Case 3-7 Project Pickings 121 Chapter Four The Art of Modeling with Spreadsheets 124 A Case Study: The Everglade Golden Years Company Cash Flow Problem 125 4.2 Overview of the Process of Modeling with Spreadsheets 126 4.3 Some Guidelines for Building “Good” Spreadsheet Models 136 4.4 Debugging a Spreadsheet Model 142 4.5 Summary 146 Glossary 146 Learning Aids for This Chapter 146 Solved Problems 146 Problems 147 Case 4-1 Prudent Provisions for Pensions 150 4.1 Chapter Five What-If Analysis for Linear Programming 151 The Importance of What-If Analysis to Managers 152 Continuing the Wyndor Case Study 154 The Effect of Changes in One Objective Function Coefficient 156 5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 162 5.5 The Effect of Single Changes in a Constraint 169 5.6 The Effect of Simultaneous Changes in the Constraints 175 5.7 Robust Optimization 179 5.8 Chance Constraints with Analytic Solver 182 5.9 Summary 186 Glossary 186 Learning Aids for This Chapter 187 Solved Problem 187 Problems 188 Case 5-1 Selling Soap 197 Case 5-2 Controlling Air Pollution 198 Case 5-3 Farm Management 200 Case 5-4 Assigning Students to Schools (Revisited) 202 5.1 5.2 5.3 Supplement to Chapter 5: Reduced Costs (This ­supplement is available at www.mhhe.com/Hillier6e). xv xvi Contents Chapter Six Network Optimization Problems 203 Minimum-Cost Flow Problems 204 A Case Study: The BMZ Co. Maximum Flow Problem 212 6.3 Maximum Flow Problems 215 6.4 Shortest Path Problems 219 6.5 Summary 229 Glossary 229 Learning Aids for This Chapter 230 Solved Problems 230 Problems 231 Case 6-1 Aiding Allies 235 Case 6-2 Money in Motion 238 Case 6-3 Airline Scheduling 240 Case 6-4 Broadcasting the Olympic Games 241 6.1 6.2 Supplement to Chapter 6: Minimum Spanning-Tree Problems (This supplement is available at www.mhhe.com /Hillier6e). Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions 243 A Case Study: The California Manufacturing Co. Problem 244 7.2 Using BIP for Project Selection: The Tazer Corp. Problem 251 7.3 Using BIP for The Selection of Sites for Emergency Services Facilities: The Caliente City Problem 253 7.4 Using BIP for Crew Scheduling: The Southwestern Airways Problem 257 7.5 Using Mixed BIP to Deal with Setup Costs for Initiating Production: The Revised Wyndor Problem 261 7.6 Summary 266 Glossary 266 Learning Aids for This Chapter 266 Solved Problems 266 Problems 268 Case 7-1 Assigning Art 273 Case 7-2 Stocking Sets 275 Case 7-3 Assigning Students to Schools (Revisited) 278 Case 7-4 Broadcasting the Olympic Games (Revisited) 278 7.1 Supplement 1 to Chapter 7: Advanced Formulation Techniques for Binary Integer Programming Supplement 2 to Chapter 7: Some Perspectives on Solving Binary Integer Programming Problems (These supplements are available at www.mhhe.com/Hillier6e.) Chapter Eight Nonlinear Programming 279 8.1 8.2 8.3 8.4 8.5 The Challenges of Nonlinear Programming 281 Nonlinear Programming with Decreasing ­Marginal Returns 289 Separable Programming 299 Difficult Nonlinear Programming Problems 309 Evolutionary Solver and Genetic Algorithms 310 Using Analytic Solver to Analyze a Model and Choose a Solving Method 318 8.7 Summary 322 Glossary 323 Learning Aids for This Chapter 323 Solved Problem 324 Problems 324 Case 8-1 Continuation of the Super Grain Case Study 329 Case 8-2 Savvy Stock Selection 330 Case 8-3 International Investments 331 8.6 Chapter Nine Decision Analysis 334 A Case Study: The Goferbroke Company Problem 335 9.2 Decision Criteria 337 9.3 Decision Trees 342 9.4 Sensitivity Analysis with Decision Trees 346 9.5 Checking Whether to Obtain More Information 350 9.6 Using New Information to Update the Probabilities 353 9.7 Using a Decision Tree to Analyze the Problem with a Sequence of Decisions 357 9.8 Performing Sensitivity Analysis on the Problem with a Sequence of Decisions 364 9.9 Using Utilities to Better Reflect the Values of Payoffs 367 9.10 The Practical Application of Decision Analysis 378 9.11 Summary 379 Glossary 379 Learning Aids for This Chapter 380 Solved Problems 381 Problems 381 Case 9-1 Who Wants to Be a Millionaire? 391 Case 9-2 University Toys and the Business Professor Action Figures 392 Case 9-3 Brainy Business 393 Case 9-4 Smart Steering Support 394 9.1 Supplement 1 to Chapter 9: Decision Criteria Supplement 2 to Chapter 9: Using TreePlan Software for Decision Trees (These supplements are available at www .mhhe.com/Hillier6e.) Chapter Ten Forecasting 397 An Overview of Forecasting Techniques 398 A Case Study: The Computer Club Warehouse (CCW) Problem 400 10.3 Applying Time-Series Forecasting Methods to the Case Study 404 10.4 The Time-Series Forecasting Methods in Perspective 423 10.5 Causal Forecasting with Linear Regression 426 10.6 Judgmental Forecasting Methods 431 10.7 Summary 433 Glossary 433 Summary of Key Formulas 434 10.1 10.2 Contents xvii Learning Aids for This Chapter 435 Solved Problem 435 Problems 435 Case 10-1 Finagling the Forecasts 442 Chapter Eleven Queueing Models 446 Elements of a Queueing Model 447 Some Examples of Queueing Systems 453 Measures of Performance for Queueing Systems 455 11.4 A Case Study: The Dupit Corp. Problem 458 11.5 Some Single-Server Queueing Models 461 11.6 Some Multiple-Server Queueing Models 469 11.7 Priority Queueing Models 474 11.8 Some Insights about Designing Queueing Systems 480 11.9 Economic Analysis of the Number of Servers to Provide 485 11.10 Summary 488 Glossary 489 Key Symbols 490 Learning Aids for This Chapter 490 Solved Problem 490 Problems 491 Case 11-1 Queueing Quandary 497 Case 11-2 Reducing In-Process Inventory 498 11.1 11.2 11.3 Supplement to Chapter 11: Additional Queueing Models (This supplement is available at www.mhhe.com/Hillier6e.) Chapter Twelve Computer Simulation: Basic Concepts 499 The Essence of Computer Simulation 500 A Case Study: Herr Cutter’s Barber Shop (Revisited) 512 12.3 Analysis of the Case Study 519 12.4 Outline of a Major Computer Simulation Study 526 12.5 Summary 529 Glossary 529 12.1 12.2 Learning Aids for This Chapter 530 Solved Problem 530 Problems 530 Case 12-1 Planning Planers 534 Case 12-2 Reducing In-Process Inventory (Revisited) 535 Supplement to Chapter 12: The Inverse Transformation Method for Generating Random Observations (This supplement is available at www.mhhe.com/Hillier6e.) Chapter Thirteen Computer Simulation with Analytic Solver 536 A Case Study: Freddie the Newsboy’s Problem 537 Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study 547 13.3 Project Management: Revisiting the Reliable Construction Co. Case Study 551 13.4 Financial Risk Analysis: Revisiting the Think-Big Development Co. Problem 557 13.5 Revenue Management in the Travel Industry 562 13.6 Choosing the Right Distribution 568 13.7 Decision Making with Parameter Analysis Reports and Trend Charts 579 13.8 Optimizing with Computer Simulation Using the Solver in Analytic Solver 587 13.9 Summary 595 Glossary 596 Learning Aids for This Chapter 596 Solved Problem 596 Problems 597 Case 13-1 Action Adventures 602 Case 13-2 Pricing under Pressure 603 Case 13-3 Financial Planning for Retirement 604 13.1 13.2 Appendix A: Tips for Using Microsoft Excel for Modeling 606 Appendix B: Partial Answers to Selected Problems 612 Index 616 CHAPTERS available at www.mhhe.com/Hillier6e Chapter Fourteen Solution Concepts for Linear Programming Some Key Facts about Optimal Solutions The Role of Corner Points in Searching for an Optimal Solution 14.3 Solution Concepts for the Simplex Method 14.4 The Simplex Method with Two Decision Variables 14.5 The Simplex Method with Three Decision Variables 14.6 The Role of Supplementary Variables 14.7 Some Algebraic Details for the Simplex Method 14.8 Computer Implementation of the Simplex Method 14.9 The Interior-Point Approach to Solving Linear Programming Problems 14.10 Summary 14.1 14.2 Glossary Learning Aids for This Chapter Problems Chapter Fifteen Transportation and Assignment Problems 15.1 15.2 15.3 15.4 15.5 A Case Study: The P & T Company Distribution Problem Characteristics of Transportation Problems Modeling Variants of Transportation Problems Some Other Applications of Variants of Transportation Problems A Case Study: The Texago Corp. Site Selection Problem xviii Contents 15.6 Characteristics of Assignment Problems 15.7 Modeling Variants of Assignment Problems 15.8 Summary Glossary Learning Aids for This Chapter Problems Case 15-1 Continuation of the Texago Case Study 18.7 The EOQ Model with Gradual Replenishment 18.8 Summary Glossary Learning Aids for This Chapter Problems Case 18-1 Brushing Up on Inventory Control Chapter Sixteen Inventory Management with Uncertain Demand PERT/CPM Models for Project Management A Case Study: The Reliable Construction Co. Project Using a Network to Visually Display a Project Scheduling a Project with PERT/CPM Dealing with Uncertain Activity Durations Considering Time–Cost Trade-Offs Scheduling and Controlling Project Costs An Evaluation of PERT/CPM from a Managerial Perspective 16.8 Summary Glossary Learning Aids for This Chapter Problems Case 16-1 Steps to Success Case 16-2 “School’s Out Forever . . .” 16.1 16.2 16.3 16.4 16.5 16.6 16.7 Chapter Seventeen Goal Programming A Case Study: The Dewright Co. Goal-Programming Problem 17.2 Weighted Goal Programming 17.3 Preemptive Goal Programming 17.4 Summary Glossary Learning Aids for This Chapter Problems Case 17-1 A Cure for Cuba Case 17-2 Remembering September 11 17.1 Chapter Eighteen Inventory Management with Known Demand 18.1 18.2 18.3 18.4 18.5 18.6 A Case Study: The Atlantic Coast Tire Corp. (ACT) Problem Cost Components of Inventory Models The Basic Economic Order Quantity (EOQ) Model The Optimal Inventory Policy for the Basic EOQ Model The EOQ Model with Planned Shortages The EOQ Model with Quantity Discounts Chapter Ninteen A Case Study for Perishable Products: Freddie the Newsboy’s Problem 19.2 An Inventory Model for Perishable Products 19.3 A Case Study for Stable Products: The Niko Camera Corp. Problem 19.4 The Management Science Team’s Analysis of the Case Study 19.5 A Continuous-Review Inventory Model for Stable Products 19.6 Larger Inventory Systems in Practice 19.7 Summary Glossary Learning Aids for This Chapter Problems Case 19-1 TNT: Tackling Newsboy’s Teachings Case 19-2 Jettisoning Surplus Stock 19.1 Chapter Twenty Computer Simulation with Crystal Ball A Case Study: Freddie the Newsboy’s Problem Bidding for a Construction Project: A Prelude to the Reliable Construction Co. Case Study 20.3 Project Management: Revisiting the Reliable Construction Co. Case Study 20.4 Cash Flow Management: Revisiting the Everglade Golden Years Company Case Study 20.5 Financial Risk Analysis: Revisiting the Think-Big Development Co. Problem 20.6 Revenue Management in the Travel Industry 20.7 Choosing the Right Distribution 20.8 Decision Making with Decision Tables 20.9 Optimizing with OptQuest 20.10 Summary Glossary Learning Aids for This Chapter Solved Problem Problems Case 20-1 Action Adventures Case 20-2 Pricing under Pressure 20.1 20.2 Chapter One Introduction Learning Objectives After completing this chapter, you should be able to 1. Define the term management science. 2. Describe the nature of management science. 3. Explain what a mathematical model is. 4. Use a mathematical model to perform break-even analysis. 5. Use a spreadsheet model to perform break-even analysis. 6. Describe the relationship between analytics and management science. 7. Identify the levels of annual savings that management science sometimes can provide to organizations. 8. Identify some special features of this book. Welcome to the field of management science! We think that it is a particularly exciting and interesting field. Exciting because management science is having a dramatic impact on the profitability of numerous business firms around the world. Interesting because the methods used to do this are so ingenious. We are looking forward to giving you a guided tour to introduce you to the special features of the field. Some people approach management science with a certain amount of anxiety and skepticism. The main source of the anxiety is the reputation of the field as being highly mathematical. This reputation then generates skepticism that such a theoretical approach can have much relevance for dealing with practical managerial problems. Most traditional courses (and textbooks) about management science have only reinforced these perceptions by emphasizing the mathematics of the field rather than its practical application. Rest easy. This is not a traditional management science textbook. We realize that most readers of this book are aspiring to become managers, not mathematicians. Therefore, the emphasis throughout is on conveying what a future manager needs to know about management science. Yes, this means including a little mathematics here and there, because it is a major language of the field. The mathematics you do see will be at the level of high school algebra plus (in the later chapters) basic concepts of elementary probability theory. We think you will be pleasantly surprised by the new appreciation you gain for how useful and intuitive mathematics at this level can be. However, managers do not need to know any of the heavy mathematical theory that underlies the various techniques of management science. Therefore, the use of mathematics plays only a strictly secondary role in the book. One reason we can deemphasize mathematics is that powerful spreadsheet software is available for applying management science. Spreadsheets provide a comfortable and familiar environment for formulating and analyzing managerial problems. The spreadsheet takes care of applying the necessary mathematics automatically in the background with only a minimum of guidance by the user. This has revolutionized the use of management science. In the past, technically trained management scientists were needed to carry out significant management science studies for management. Now spreadsheets are bringing many of the tools and concepts of management science within the reach of managers for conducting their own analyses. Although busy managers will continue to call upon management science teams to conduct major studies for them, they are increasingly becoming direct users themselves through the 1 2 Chapter One Introduction medium of spreadsheet software. Therefore, since this book is aimed at future managers (and management consultants), we will emphasize the use of spreadsheets for applying management science. What does an enlightened future manager need to learn from a management science course? 1. Gain an appreciation for the relevance and power of management science. (Therefore, we include many application vignettes throughout the book that give examples of actual applications of management science and the impact they had on the organizations involved.) 2. Learn to recognize when management science can (and cannot) be fruitfully applied. (Therefore, we will emphasize the kinds of problems to which the various management science techniques can be applied.) 3. Learn how to apply the major techniques of management science to analyze a variety of managerial problems. (Therefore, we will focus largely on how spreadsheets enable many such applications with no more background in management science than provided by this book.) 4. Develop an understanding of how to interpret the results of a management science study. (Therefore, we will present many case studies that illustrate management science studies and how their results depend on the assumptions and data that were used.) The objectives just described are the key teaching goals of this book. We begin this process in the next four sections by introducing the nature of management science and the impact that it is having on many organizations. (This process will continue throughout the remaining chapters as well.) Section 1.5 then points out some of the special features of this book that you can look forward to seeing in the subsequent chapters. 1.1 THE NATURE OF MANAGEMENT SCIENCE What is the name management science (sometimes abbreviated MS) supposed to convey? It does involve management and science or, more precisely, the science of management, but this still is too vague. Here is a more suggestive definition. Management science is a discipline that attempts to aid managerial decision making by applying a scientific approach to managerial problems that involve quantitative factors. Now let us see how elaborating upon each of the italicized terms in this definition conveys much more about the nature of management science. Management Science Is a Discipline As a discipline, management science is a whole body of knowledge and techniques that are based on a scientific foundation. For example, it is analogous in some ways to the medical field. A medical doctor has been trained in a whole body of knowledge and techniques that are based on the scientific foundations of the medical field. After receiving this training and entering practice, the doctor must diagnose a patient’s illness and then choose the appropriate medical procedures to apply to the illness. The patient then makes the final decision on which medical procedures to accept. For less serious cases, the patient may choose not to consult a doctor and instead use his own basic knowledge of medical principles to treat himself. Similarly, a management scientist must receive substantial training (albeit considerably less than for a medical doctor). This training also is in a whole body of knowledge and techniques that are based on the scientific foundations of the discipline. After entering practice, the management scientist must diagnose a managerial problem and then choose the appropriate management science techniques to apply in analyzing the problem. The cognizant manager then makes the final decision as to which conclusions from this analysis to accept. For less extensive managerial problems where management science can be helpful, the manager may choose not to consult a management scientist and instead use his or her own basic knowledge of management science principles to analyze the problem. Although it has considerably longer roots, the rapid development of the discipline began in the 1940s and 1950s. The initial impetus came early in World War II, when large numbers of scientists were called upon to apply a scientific approach to the management of the war effort 1.1 The Nature of Management Science 3 operations research Management science began its rapid development during World War II with the name operations research. for the allies. When the war ended, the success of this approach in the war effort spurred interest in applying it outside the military as well. By the early 1950s, substantial applications of management science were being seen in a variety of organizations in business, industry, and government. Another landmark event in the history of management science was the discovery in 1947 by George Dantzig of the simplex method for solving linear programming problems. (Linear programming is the subject of several early chapters.) Considerable progress in developing the other techniques of management science also occurred throughout the middle of the 20th century. However, the very limited computational power available at that time (whether when doing the computations by hand or with the relatively primitive electronic computers of the day) prevented applying these techniques except to small problems. Fortunately, another factor that gave great impetus to the growth of the discipline ever since that time was the onslaught of the computer revolution. Even massive problems usually can be solved now with today’s powerful computers. The traditional name given to the discipline (and the one that still is widely used today outside of business schools) is operations research. This name was applied because the teams of scientists in World War II were doing research on how to manage military operations. The abbreviation OR also is widely used. This abbreviation often is combined with the one for management science (MS), thereby referring to the discipline as OR/MS. According to estimates from the U.S. Bureau of Labor Statistics in 2015, there were approximately 96,000 individuals at that time working as operations research analysts in the United States (some with just a B.S. degree) with an average annual salary of about $84,000. The Bureau also forecasted that this number of individuals working as operations research analysts would grow by 30 percent over the subsequent decade. Another discipline that is closely related to management science is analytics (sometimes called business analytics when dealing with business problems). Like management science, analytics attempts to aid managerial decision making but with particular emphasis on three types of analysis: (1) descriptive analytics—the use of data (sometimes massive amounts of data) to analyze trends, (2) predictive analytics—the use of data to predict what will happen in the future (perhaps by using the forecasting techniques described in Chapter 10), and (3) prescriptive analytics—the use of data to prescribe the best course of action (frequently by using the optimization techniques described throughout this book). Broadly speaking, the techniques of the management science discipline provide the firepower for prescriptive analytics and, to a lesser extent, for predictive analytics, but not so much for descriptive analytics. (Section 1.3 will further describe the relationship between analytics and management science.) One major international professional society for the management science discipline (as well as for business analytics) is the Institute for Operations Research and the Management Sciences (INFORMS). Headquartered in the United States, with well over 12,000 members, this society holds major conferences in the United States each year (including an annual Conference for Business Analytics and Operations Research) plus occasional conferences elsewhere. It also publishes several prominent journals, including Management Science, Operations Research, Analytics, and Interfaces. (Articles describing actual applications of management science are featured in Interfaces, so you will see many references and links to this journal throughout the book.) In addition, a few dozen countries around the world have their own national operations research societies. (More about this in Section 1.4.) Thus, operations research/management science (OR/MS) is a truly international discipline. (We hereafter will normally just use the name management science or the abbreviation MS.) Management Science Aids Managerial Decision Making The key word here is that management science aids managerial decision making. Management scientists don’t make managerial decisions. Managers do. A management science study only provides an analysis and recommendations, based on the quantitative factors involved in the problem, as input to the cognizant managers. Managers must also take into account various intangible considerations that are outside the realm of management science and then use their best judgment to make the decision. Sometimes managers find that qualitative factors are as important as quantitative factors in making a decision. 4 Chapter One Introduction A small informal management science study might be conducted by just a single individual, who may be the cognizant manager. However, management science teams normally are used for larger studies. (We often will use the term team to cover both cases throughout the book.) Such a team often includes some members who are not management scientists but who provide other types of expertise needed for the study. Although a management science team often is entirely in-house (employees of the company), part or all of the team may instead be consultants who have been hired for just the one study. Consulting firms that partially or entirely specialize in management science currently are a growing industry. Management Science Uses a Scientific Approach Management science is based strongly on some scientific fields, including mathematics, statistics, and computer science. It also draws on the social sciences, especially economics. Since the field is concerned with the practical management of organizations, a management scientist should have solid training in business administration, including its various functional areas, as well. To a considerable extent, a management science team will attempt to use the scientific method in conducting its study. This means that the team will emphasize conducting a systematic investigation that includes careful data gathering, developing and testing hypotheses about the problem (typically in the form of a mathematical model), and then applying sound logic in the subsequent analysis. When conducting this systematic investigation, the management science team typically will follow the (overlapping) steps outlined and described below. Step 1: Define the problem and gather data. In this step, the team consults with management to clearly identify the problem of concern and ascertain the appropriate objectives for the study. The team then typically spends a surprisingly large amount of time gathering relevant data about the problem with the assistance of other key individuals in the organization. A common frustration is that some key data are either very rough or completely unavailable. This may necessitate installing a new computer-based management information system. Another increasingly common problem is that there may be too much data available to be easily analyzed. Dramatic advances in computerized data capture, processing power, data transmission, and storage capabilities are enabling organizations to integrate their various databases into massive data warehouses. This has led to the development of datamining software for extracting hidden predictive information, correlations, and patterns from large databases. Fortunately, the rapid development of the information technology (IT) field in recent years is leading to a dramatic improvement in the quantity and quality of data that may be available to the management science (MS) team. Corporate IT now is often able to provide the computational resources and databases, as well as any helpful data mining, that are needed by the MS team. Thus, the MS team often will collaborate closely with the IT group. Step 2: Formulate a model (typically a mathematical model) to represent the problem. Models, or approximate representations, are an integral part of everyday life. Common examples include model airplanes, portraits, globes, and so on. Similarly, models play an important role in science and business, as illustrated by models of the atom, models of genetic structure, mathematical equations describing physical laws of motion or chemical reactions, graphs, organization charts, and industrial accounting systems. Such models are invaluable for abstracting the essence of the subject of inquiry, showing interrelationships, and facilitating analysis. Mathematical models are also approximate representations, but they are expressed in terms of mathematical symbols and expressions. Such laws of physics as F = ma and E = mc2 are familiar examples. Similarly, the mathematical model of a business problem is the system of equations and related mathematical expressions that describes the essence of the problem. With the emergence of powerful spreadsheet technology, spreadsheet models now are widely used to analyze managerial problems. A spreadsheet model lays out the relevant data, measures of performance, interrelationships, and so forth, on a spreadsheet in 1.1 The Nature of Management Science 5 an organized way that facilitates fruitful analysis of the problem. It also frequently incorporates an underlying mathematical model to assist in the analysis, but the mathematics is kept in the background so the user can concentrate on the analysis. The modeling process is a creative one. When dealing with real managerial problems (as opposed to some cut-and-dried textbook problems), there normally is no single “correct” model but rather a number of alternative ways to approach the problem. The modeling process also is typically an evolutionary process that begins with a simple “verbal model” to define the essence of the problem and then gradually evolves into increasingly more complete mathematical models (perhaps in a spreadsheet format). We further describe and illustrate such mathematical models in the next section. Step 3: Develop a computer-based procedure for deriving solutions to the problem from the model. The beauty of a well-designed mathematical model is that it enables the use of mathematical procedures to find good solutions to the problem. These procedures usually are run on a computer because the calculations are too extensive to be done by hand. In some cases, the management science team will need to develop the procedure. In others, a standard software package already will be available for solving the model. When the mathematical model is incorporated into a spreadsheet, the spreadsheet software normally includes a Solver that usually will solve the model. Step 4: Test the model and refine it as needed. Now that the model can be solved, the team needs to thoroughly check and test the model to make sure that it provides a sufficiently accurate representation of the real problem. A number of questions should be addressed, perhaps with the help of others who are particularly familiar with the problem. Have all the relevant factors and interrelationships in the problem been accurately incorporated into the model? Does the model seem to provide reasonable solutions? When it is applied to a past situation, does the solution improve upon what was actually done? When assumptions about costs and revenues are changed, do the solutions change in a plausible manner? Step 5: Apply the model to analyze the problem and develop recommendations for management. The management science team now is ready to solve the model, perhaps under a variety of assumptions, in order to analyze the problem. The resulting recommendations then are presented to the managers who must make the decisions about how to deal with the problem. If the model is to be applied repeatedly to help guide decisions on an ongoing basis, the team might also develop a decision support system. This is an interactive ­computer-based system that aids managerial decision making. The system draws current data from databases or management information systems and then solves the various ­versions of the model specified by the manager. Step 6: Help to implement the team’s recommendations that are adopted by management. Once management makes its decisions, the management science team normally is asked to help oversee the implementation of the new procedures. This includes providing some information to the operating management and personnel involved on the rationale for the changes that are being made. The team also makes sure that the new operating system is consistent with its recommendations as they have been modified and approved by management. If successful, the new system may be used for years to come. With this in mind, the team monitors the initial experience with the system and seeks to identify any modifications that should be made in the future. Management Science Considers Quantitative Factors Many managerial problems revolve around such quantitative factors as production quantities, revenues, costs, the amounts available of needed resources, and so on. By incorporating these quantitative factors into a mathematical model and then applying mathematical procedures to solve the model, management science provides a uniquely powerful way of analyzing such managerial problems. Although management science is concerned with the practical management of organizations, including taking into account relevant qualitative factors, its special contribution lies in this unique ability to deal with the quantitative factors. 6 Chapter One Introduction The Special Products Company example discussed next will illustrate how management science considers quantitative factors. Review Questions 1. When did the rapid development of the management science discipline begin? 2. What is the traditional name given to this discipline that still is widely used outside of business schools? 3. What does a management science study provide to managers to aid their decision making? 4. Upon which scientific fields and social sciences is management science especially based? 5. What is a decision support system? 6. What are some common quantitative factors around which many managerial problems revolve? 1.2 AN ILLUSTRATION OF THE MANAGEMENT SCIENCE APPROACH: BREAK-EVEN ANALYSIS A cost that remains the same regardless of the production volume is referred to as a fixed cost, whereas a cost that varies with the production volume is called a variable cost. The Special Products Company produces expensive and unusual gifts to be sold in stores that cater to affluent customers who already have everything. The latest new-product proposal to management from the company’s Research Department is a first-of-its-kind iWatch. This iWatch would combine the features of a top-of-the-line atomic wristwatch and a next-­ generation smartphone, including the ability to respond to voice commands or questions with voice responses. It also would connect to the Internet wirelessly to provide weather, sports scores, stock quotes, and more. An extensive research-and-development project would be needed to develop the iWatch. The proposal is to provide a generous budget of $10 million for this project in order to provide as many desirable features as possible within this budget. It is clear that the production costs for the iWatch would be very large because of the extreme miniaturization that would be required, so the selling price would need to be far beyond the reach of middle-class customers. Therefore, the marketing of the iWatch would be aimed at wealthy customers who want the most advanced products regardless of cost. Management needs to decide whether to develop and market this new product and, if so, how many of these watches to produce. Before making these decisions, a sales forecast will be obtained to estimate how many watches can be sold. Since most of these sales would occur quickly during the relatively brief time before the “next big thing” arrives to take over the market, there would be only one production run for the iWatch and the number produced would be set equal to the sales forecast. Following the production run, the iWatch would be marketed as aggressively as needed to sell this entire inventory if possible. Management now needs a management science study to be conducted to determine how large this sales potential needs to be to make the iWatch profitable after considering all the prospective revenues and costs, so let’s next look at the estimates of these financial figures. If the company goes ahead with this product, the research-and-development cost of $10 million is referred to as a fixed cost because it remains the same regardless of how many watches are produced and sold. (However, note that this cost would not be incurred if management decides not to introduce the product since the research-and-development project then would not be undertaken.) In addition to this fixed cost, there is a production cost that varies with the number of watches produced. This variable cost is $1,000 per watch produced, which adds up to $1,000 times the number of watches produced. (The cost for each additional unit produced, $1,000, is referred to as the marginal cost.) Each watch sold would generate a unit revenue of $2,000 for the company. Spreadsheet Modeling of the Problem You will see throughout this book that spreadsheets provide a very convenient way of using a management science approach for modeling and analyzing a wide variety of managerial problems. This certainly is true for the Special Products Company problem as well, as we now will demonstrate. Figure 1.1 shows a spreadsheet formulation of this problem after obtaining a preliminary sales forecast that indicates 30,000 watches can be sold. The data have been entered into cells C4 to C7. Cell C9 is used to record a trial value for the decision as to how many watches to produce. As one of the many possibilities that eventually might be tried, Figure 1.1 shows the specific trial value of 20,000. 1.2 An Illustration of the Management Science Approach: Break-Even Analysis 7 FIGURE 1.1 A spreadsheet formulation of the Special Products Company problem. A 1 B C Special Products Co. Break-Even Analysis D E F 2 3 Data 4 5 Unit Revenue Fixed Cost $2,000 $10,000,000 6 Marginal Cost 7 Results Total Revenue Total Fixed Cost $40,000,000 $10,000,000 $1,000 Total Variable Cost $20,000,000 Sales Forecast 30,000 Profit (Loss) $10,000,000 Production Quantity 20,000 8 9 Range Name FixedCost MarginalCost ProductionQuantity Profit SalesForecast TotalFixedCost TotalRevenue TotalVariableCost UnitRevenue Excel Tip: To update formulas throughout the spreadsheet to incorporate a newly defined range name, choose Apply Names from the Define Name menu on the Formulas tab. Excel Tip: A list of all the defined names and their corresponding cell references can be pasted into a spreadsheet by choosing Paste Names from the Use in Formula menu on the Formulas tab, and then clicking on Paste List. A spreadsheet is a convenient tool for performing what-if analysis. The Excel function MIN (a, b) gives the minimum of the numbers in the cells whose addresses are a and b. Cell C5 C6 C9 F7 C7 F5 F4 F6 C4 E 3 4 5 6 7 F Results Total Revenue =UnitRevenue * MIN(SalesForecast, ProductionQuantity) Total Fixed Cost =IF(ProductionQuantity > 0, FixedCost, 0) Total Variable Cost =MarginalCost * ProductionQuantity Profit (Loss) =TotalRevenue – (TotalFixedCost + TotalVariableCost) Cells F4 to F7 give the resulting total revenue, total costs, and profit (loss) by using the Excel equations shown under the spreadsheet in Figure 1.1. The Excel equations could have been written using cell references (e.g., F6 = C6*C9). However, the spreadsheet model is made clearer by giving “range names” to key cells or blocks of cells. (A range name is a descriptive name given to a cell or range of cells that immediately identifies what is there. Appendix A provides details about how to incorporate range names into a spreadsheet model.) To define a name for a selected cell (or range of cells), click on the name box (on the left of the formula bar above the spreadsheet) and type a name. These cell names then can be used in other formulas to create an equation that is easy to decipher (e.g., TotalVariableCost = MarginalCost*ProductionQuantity rather than the more cryptic F6 = C6*C9). Note that spaces are not allowed in range names. When a range name has more than one word, we have used capital letters to distinguish the start of each new word (e.g., ProductionQuantity). The lower left-hand corner of Figure 1.1 lists the names of the quantities in the spreadsheet in alphabetical order and then gives cell references where the quantities are found. Although this isn’t particularly necessary for such a small spreadsheet, you should find it helpful for the larger spreadsheets found later in the book. This same spreadsheet, along with all of the other spreadsheets in the book, are available to you at www.mhhe.com/Hillier6e. As you can see for yourself by bringing up and playing with the spreadsheet, it provides a straightforward way of performing what-if analysis on the problem. What-if analysis involves addressing such questions as, What happens if the sales forecast should have been considerably lower? What happens if some of the cost and revenue estimates are wrong? Simply enter a variety of new values for these quantities in the spreadsheet and see what happens to the profit shown in cell F7. The lower right-hand corner of Figure 1.1 introduces two useful Excel functions, the MIN(a, b) function and the IF(a, b, c) function. The equation for cell F4 uses the MIN(a, b) function, which gives the minimum of a and b. In this case, the estimated number of watches that will be sold is the minimum of the sales forecast and the production quantity, so F4 = UnitRevenue*MIN(SalesForecast, ProductionQuantity) 8 Chapter One Introduction enters the unit revenue (from cell C4) times the minimum of the sales forecast (from C7) and the production quantity (from C9) into cell F4. Also note that the equation for cell F5 uses the IF(a, b, c) function, which does the following: If statement a is true, it uses b; otherwise, it uses c. Therefore, The Excel function IF (a, b, c) tests if a is true. If so, it uses b; otherwise it uses c. F5 = IF(ProductionQuantity > 0, FixedCost, 0) says to enter the fixed cost (C5) into cell F5 if the production quantity (C9) is greater than zero, but otherwise enter 0 (the fixed cost is avoided if production is not initiated). The spreadsheet in Figure 1.1, along with its equations for the results in column F, constitutes a spreadsheet model for the Special Products Company problem. You will see many examples of such spreadsheet models throughout the book. This particular spreadsheet model is based on an underlying mathematical model that uses algebra to spell out the equations in cells F4:F7 and then to derive some additional useful information. Let us take a look at this mathematical model next. Expressing the Problem Mathematically The issue facing management is to make the following decision. Decision to be made: Number of watches to produce (if any). Since this number is not yet known, we introduce an algebraic variable Q to represent this quantity. Thus, Q = Number of watches to produce, where Q is referred to as a decision variable. Naturally, the value chosen for Q should not exceed the sales forecast for the number of watches that can be sold. Choosing a value of 0 for Q would correspond to deciding not to introduce the product, in which case none of the costs or revenues described in the preceding paragraph would be incurred. The objective is to choose the value of Q that maximizes the company’s profit from this new product. The management science approach is to formulate a mathematical model to represent this problem by developing an equation that expresses the profit in terms of the decision variable Q. To get there, it is necessary first to develop equations in terms of Q for the total cost and revenue generated by the watches. If Q = 0, no cost is incurred. However, if Q > 0, there is both a fixed cost and a variable cost. Fixed cost = $10 million (if Q > 0) Variable cost = $1,000Q Therefore, the total cost would be 0 if Q = 0 Total cost = {$10 million + $1,000Q if Q > 0 Since each watch sold would generate a revenue of $2,000 for the company, the total revenue from selling Q watches would be Total revenue = $2,000Q Consequently, the profit from producing and selling Q watches would be Profit = Total revenue − Total cost if Q = 0 0 = {$2,000Q − ($10 million + $1,000Q) if Q > 0 Thus, since $2,000Q − $1,000Q = $1,000Q Profit = −$10 million + $1,000Q if Q > 0 Analysis of the Problem This last equation shows that the attractiveness of the proposed new product depends greatly on the value of Q, that is, on the number of watches that can be produced and sold. A small 1.2 An Illustration of the Management Science Approach: Break-Even Analysis 9 FIGURE 1.2 Break-even analysis for the Special Products Company shows that the cost line and revenue line intersect at Q = 10,000 watches, so this is the break-even point for the proposed new product. Revenue/Cost in $millions 50 40 Profit Total revenue = $2000Q 30 20 10 0 Total cost = $10 million + $1000Q if Q > 0 Fixed cost = $10 million if Q > 0 Loss 5 10 15 20 25 Q Production Quantity (thousands) Break-even point = 10,000 watches value of Q means a loss (negative profit) for the company, whereas a sufficiently large value would generate a positive profit for the company. For example, look at the difference between Q = 2,000 and Q = 20,000. Profit = − $10 million + $1,000 (2,000) = −$8 million if Q = 2,000 Profit = − $10 million + $1,000 (20,000) = $10 million if Q = 20,000 Figure 1.2 plots both the company’s total cost and total revenue for the various values of Q. Note that the cost line and the revenue line intersect at Q = 10,000. For any value of Q < 10,000, cost exceeds revenue, so the gap between the two lines represents the loss to the company. For any Q > 10,000, revenue exceeds cost, so the gap between the two lines now shows positive profit. At Q = 10,000, the profit is 0. Since 10,000 units is the production and sales volume at which the company would break even on the proposed new product, this volume is referred to as the break-even point. This is the point that must be exceeded to make it worthwhile to introduce the product. Therefore, the crucial question is whether the sales forecast for how many watches can be sold is above or below the break-even point. Figure 1.2 illustrates the graphical procedure for finding the break-even point. Another alternative is to use an algebraic procedure to solve for the point. Because the profit is 0 at this point, the procedure consists of solving the following equation for the unknown Q. Profit = −$10 million + $1,000Q = 0 Thus, $1,000Q = $10 million $10 million = __________ Q $1,000 Q = 10,000 A Complete Mathematical Model for the Problem The preceding analysis of the problem made use of a basic mathematical model that consisted of the equation for profit expressed in terms of Q. However, implicit in this analysis were some additional factors that can be incorporated into a complete mathematical model for the problem. 10 Chapter One Introduction Two of these factors concern restrictions on the values of Q that can be considered. One of these is that the number of watches produced cannot be less than 0. Therefore, Q ≥ 0 constraints A constraint in a mathematical model is an inequality or equation that expresses some restrictions on the values that can be assigned to the decision variables. is one of the constraints for the complete mathematical model. Another restriction on the value of Q is that it should not exceed the number of watches that can be sold. The preliminary sales forecast is 30,000, but this is such a crucial number that more time is needed now to develop a final sales forecast that will be used hereafter. This final sales forecast has not yet been obtained, so let the symbol s represent this currently unknown value. s = Sales forecast (not yet available) of the number of watches that can be sold Consequently, Q ≤ s parameter The constants in a mathematical model are referred to as the parameters of the model. is another constraint, where s is a parameter of the model whose value has not yet been chosen. The final factor that should be made explicit in the model is the fact that management’s objective is to make the decision that maximizes the company’s profit from this new product. Therefore, the complete mathematical model for this problem is to find the value of the decision variable Q so as to 0 if Q = 0 Maximize profit = {− $10 million + $1,000Q if Q > 0 subject to objective function The objective function for a mathematical model is a mathematical expression that gives the measure of performance for the problem in terms of the decision variables. Q≤s Q≥0 where the algebraic expression given for Profit is called the objective function for the model. The value of Q that solves this model depends on the value that will be assigned to the parameter s (the future final forecast of the number of units that can be sold). Because the break-even point is 10,000, here is how the solution for Q depends on s. Solution for Mathematical Model: Fixed cost $10 million _________________________ Break-even point = = ______________ − $1,000 = 10,000 Unit revenue − Marginal cost $2,000 If s ≤ 10,000, then set Q = 0. If s >10,000, then set Q = s. Therefore, the company should introduce the product and produce the number of units that can be sold only if this production and sales volume exceeds the break-even point. What-if Analysis of the Mathematical Model what-if analysis Since estimates can be wrong, what-if analysis is used to check the effect on the recommendations of a model if the estimates turn out to be wrong. A mathematical model is intended to be only an approximate representation of the problem. For example, some of the numbers in the model inevitably are only estimates of quantities that cannot be determined precisely at this time. The above mathematical model is based on four numbers that are only estimates—the fixed cost of $10 million, the marginal cost of $1,000, the unit revenue of $2,000, and the final sales forecast (after it is obtained). A management science study usually devotes considerable time to investigating what happens to the recommendations of the model if any of the estimates turn out to considerably miss their targets. This is referred to as what-if analysis. Incorporating the Break-Even Point into the Spreadsheet Model A key finding of the above mathematical model is its formula for the break-even point, Fixed cost Break-even point = _________________________ Unit revenue − Marginal cost Therefore, once both the quantities in this formula and the sales forecast have been carefully estimated, the solution for the mathematical model specifies what the production quantity should be. By contrast, although the spreadsheet in Figure 1.1 enables trying a variety of trial values for the production quantity, it does not directly indicate what the production quantity 1.2 An Illustration of the Management Science Approach: Break-Even Analysis 11 FIGURE 1.3 An expansion of the spreadsheet in Figure 1.1 that uses the solution for the mathematical model to calculate the break-even point. A 1 B C Special Products Co. Break-Even Analysis D E F 2 3 Data 4 5 Unit Revenue Fixed Cost $2,000 $10,000,000 6 Marginal Cost 7 Results Total Revenue Total Fixed Cost $60,000,000 $10,000,000 $1,000 Total Variable Cost $30,000,000 Sales Forecast 30,000 Profit (Loss) $20,000,000 Production Quantity 30,000 Break-Even Point 8 9 Range Name BreakEvenPoint FixedCost MarginalCost ProductionQuantity Profit SalesForecast TotalFixedCost TotalRevenue TotalVariableCost UnitRevenue Cell F9 C5 C6 C9 F7 C7 F5 F4 F6 C4 E 3 4 5 6 7 10,000 F Results Total Revenue =UnitRevenue * MIN(SalesForecast, ProductionQuantity) Total Fixed Cost =IF(ProductionQuantity > 0, FixedCost, 0) Total Variable Cost =MarginalCost * ProductionQuantity Profit (Loss) =TotalRevenue – (TotalFixedCost + TotalVariableCost) 8 9 Break-Even Point =FixedCost/(UnitRevenue – MarginalCost) should be. Figure 1.3 shows how this spreadsheet can be expanded to provide this additional guidance. Suppose that a final sales forecast of 30,000 (unchanged from the preliminary sales forecast) now has been obtained, as shown in cell C7. As indicated by its equation at the bottom of the figure, cell F9 calculates the break-even point by dividing the fixed cost ($10 million) by the net profit per watch sold ($1,000), where this net profit is the unit revenue ($2,000) minus the marginal cost ($1,000). Since the sales forecast of 30,000 exceeds the break-even point of 10,000, this forecast has been entered into cell C9. If desired, the complete mathematical model for break-even analysis can be fully incorporated into the spreadsheet by requiring that the model solution for the production quantity be entered into cell C9. This would be done by using the equation C9 = IF(SalesForecast > BreakEvenPoint, SalesForecast, 0) However, the disadvantage of introducing this equation is that it would eliminate the possibility of trying other production quantities that might still be of interest. For example, if management does not have much confidence in the final sales forecast and wants to minimize the danger of producing more watches than can be sold, consideration would be given to production quantities smaller than the forecast. For example, the trial value shown in cell C9 of Figure 1.1 might be chosen instead. As in any application of management science, a mathematical model can provide useful guidance but management needs to make the final decision after considering factors that may not be included in the model. Review Questions 1. How do the production and sales volume of a new product need to compare to its break-even point to make it worthwhile to introduce the product? 2. What are the factors included in the complete mathematical model for the Special Products Company problem, in addition to an equation for profit? 3. What is the purpose of what-if analysis? 4. How can a spreadsheet be used to perform what-if analysis? 5. What does the MIN(a, b) Excel function do? 6. What does the IF(a, b, c) Excel function do? 12 Chapter One Introduction 1.3 THE RELATIONSHIP BETWEEN ANALYTICS AND MANAGEMENT SCIENCE Analytics is a broad term that includes both management science and all the other quantitative decision sciences, such as mathematics, statistics, computer science, data science, industrial engineering, etc. By drawing on any of these tools to analyze the available data, analytics can be defined as the scientific process of transforming data into insight for making better decisions. Analytics encompasses all of the quantitative decision sciences. The era of big data has created new challenges that require the use of analytics. There has been great buzz throughout the business world in recent years about something called analytics (or business analytics) and the importance of incorporating analytics into managerial decision making. The primary impetus for this buzz was a series of articles and books by Thomas H. Davenport, a renowned thought-leader who has helped hundreds of companies worldwide to revitalize their business practices. He initially introduced the concept of analytics in the January 2006 issue of the Harvard Business Review with an article, “Competing on Analytics,” that now has been named as one of the ten must-read articles in that magazine’s 90-year history. This article soon was followed by two best-selling books entitled Competing on Analytics: The New Science of Winning and Analytics at Work: Smarter Decisions, Better Results. So what is analytics? In contrast to management science, analytics is not a single discipline with its own well-defined body of techniques. Analytics instead includes many disciplines, namely, all the quantitative decision sciences. Thus, any application of analytics draws on any of the quantitative decision sciences that can be most helpful in analyzing a given problem. Therefore, a company’s analytics group might include members with titles such as mathematician, statistician, computer scientist, data scientist, information technologist, business analyst, industrial engineer, management scientist, and operations research analyst. At this point, the members of a management science group might object, saying that their management science studies often draw upon these other quantitative decision sciences as well. This frequently is true, but it also is true that some applications of analytics draw mainly from certain other quantitative decision sciences instead of management science. This typically occurs when the issue being addressed is to try to gain insights from massive amounts of available data, so that data science and statistics become the key quantitative decision sciences. This kind of application is a major emphasis of analytics. The reason for this emphasis is that analytics fully recognizes that we have entered into the era of big data where massive amounts of data now are commonly available to many businesses and organizations to help guide managerial decision making. The current data surge is coming from sophisticated computer tracking of shipments, sales, suppliers, and customers, as well as e-mail, web traffic, and social networks. A primary focus of analytics is on how to make the most effective use of all these data. The application of analytics can be divided into three overlapping categories. Here are the traditional names and brief definitions of these categories: Category 1: Descriptive analytics (analyzing data to improve descriptions of what has been happening) Category 2: Predictive analytics (using predictive models to improve predictions of what is likely to happen in the future) Category 3: Prescriptive analytics (using prescriptive models, including optimization models, to improve managerial decision making) The first of these categories requires dealing with perhaps massive amounts of historical data where information technology has been used to store and access the data. Descriptive analytics then uses innovative techniques to locate the relevant data and identify the interesting patterns in order to describe and understand what is going on now. One important technique for doing this is called data mining. A software package accompanying this book, Analytic Solver, includes a tab called Data Mining for applying both data mining and predictive analytics. Because the analysis of data is the focus of the work being done in this category, data analytics is another name that is sometimes used for this category (and perhaps category 2 as well). Some analytics professionals who specialize in descriptive analytics are called data scientists. Predictive analytics involves applying predictive models to historical data and perhaps external data to predict future events or trends. The models underlying statistical forecasting methods, such as those described in Chapter 10, are often used here. Computer simulation (Chapters 12 and 13) also can be useful for demonstrating future events that can occur. Because some of the methods of predictive analytics are relatively sophisticated, this category tends to be more advanced than the first one. 1.3 The Relationship Between Analytics and Management Science 13 Prescriptive analytics uses powerful techniques drawn mainly from management science to prescribe what should be done in the future. Sports analytics is another example of various areas where analytics now is applied. Business schools now are responding to the great need to thoroughly train much larger numbers of people in business analytics. Prescriptive analytics is the final (and most advanced) category. It involves applying sophisticated models to the data to prescribe what should be done in the future. The powerful optimization models and techniques of management science described in many of the chapters of this book commonly are what are used here. The purpose is to guide managerial decision making, so the name decision analytics also could be used to describe this category. Management scientists often deal with all three of these categories, but not very much with the first one, somewhat more with the second one, and then heavily with the third one. Thus, management science can be thought of as focusing mainly on advanced analytics (predictive and prescriptive activities) whereas analytics professionals might get more involved than management scientists with the entire business process, including what precedes the first category (identifying a need) and what follows the last category (implementation). Looking to the future, the two approaches should tend to merge somewhat over time. Although analytics was initially introduced as a key tool for mainly business organizations, it also can be a powerful tool in other contexts. As one example, analytics (together with management science) played a key role in the 2012 presidential campaign in the United States. The Obama campaign management hired a multi-disciplinary team of statisticians, predictive modelers, data-mining experts, mathematicians, software programmers, and management scientists. It eventually built an entire analytics department five times as large as that of its 2008 campaign. With all this analytics input, the Obama team launched a full-scale and all-front campaign, leveraging massive amounts of data from various sources to directly micro-target potential voters and donors with tailored messages. The election had been expected to be a very close one, but the Obama “ground game” that had been propelled by descriptive and predictive analytics was given much of the credit for the clear-cut Obama win. Another famous application of analytics is described in the book Moneyball and a subsequent 2011 movie with the same name that is based on this book. They tell the true story of how the Oakland Athletics baseball team achieved great success, despite having one of the smallest budgets in the major leagues, by using various kinds of nontraditional data (referred to as saber metrics) to better evaluate the potential of players available through a trade or the draft. Although these evaluations often flew in the face of conventional baseball wisdom, both descriptive analytics and predictive analytics were being used to identify overlooked players who could greatly help the team. After witnessing the impact of analytics, many major league baseball teams now have hired analytics professionals. Some other kinds of sports teams, including professional basketball teams in the NBA (National Basketball Association) also have begun to use analytics. For example, some months after they won the 2015 NBA championship, the Golden State Warriors won the “Best Analytics Organization” award at the MIT Sloan Sports Analytics Conference in March 2016. The top management of numerous business organizations now understand the impact that this approach can have on the bottom line and they are interested in increasing the role of their analytics group in their organization. This will require many more people trained in analytics and management science. A recent study by the McKinsey Global Institute estimated that the United States could face a shortage by 2018 of 140,000 to 180,000 people with deep analytical skills. The study also estimated that employers will need an additional 1.5 million managers and analysts with the experience and expertise to use the analysis of big data to make effective and efficient decisions. Universities now are responding to this great need. There now are hundreds of business schools in the United States or abroad that have, or have committed to launch, curriculum at the undergraduate and graduate levels with degrees or certificates in business analytics. Courses that cover the material in this book would be a key component of these programs, along with courses that emphasize other areas of analytics (e.g., statistics and data mining). This creates an outstanding opportunity for students with a STEM focus. In the words of the thought leader Thomas H. Davenport (who was introduced in the first paragraph of this section), the job of an analytics professional promises to be the “sexiest job in the 21st century.” In 2016, the job site Glassdoor also named data scientist as the best job in America. Along these same lines, U.S. News annually publishes a list of the best jobs in the United States, based on a variety of factors such as salary, job satisfaction, etc. In their issue of January 27, 2016, U.S. News published that year’s list that placed the operations research profession (i.e., the management science profession) as number 2 on the list of the best business jobs in the country. Furthermore, this profession ranks very high in terms of having a high percentage of women working in the profession, with slightly more women than men in the field. 14 Chapter One Introduction Management science is at the core of advanced analytics. Review Questions 1.4 The momentum of the analytics movement is indeed continuing to grow rapidly. Because management science is at the core of advanced analytics, the usage of the powerful techniques of management science introduced in this book also should continue to grow rapidly. However, without even looking to the future, the impact of management science over past years also has been impressive, as described in the next section. 1. 2. 3. 4. 5. 6. What are some quantitative decision sciences that are included within analytics? What does descriptive analytics involve doing? What does predictive analytics involve doing? What does prescriptive analytics involve doing? In which categories of analytics does management science mainly focus? What is a major emphasis of analytics that draws mainly from certain other quantitative decision sciences instead of management science? THE IMPACT OF MANAGEMENT SCIENCE The most important applications of management science in business and industry have resulted in annual savings in the hundreds of millions of dollars. Management science also has had a major impact in the health care area, in guiding key governmental policies, and in military applications. Management science (or operations research as it is commonly called by practitioners) has had an impressive impact on improving the efficiency of numerous organizations around the world. In the process, management science has made a significant contribution to increasing the productivity of the economies of various countries. There now are a few dozen member countries in the International Federation of Operational Research Societies (IFORS), with each country having a national operations research society. Both Europe and Asia have federations of such societies to coordinate holding international conferences and publishing international journals in those continents. In addition, we described in Section 1.1 how the Institute for Operations Research and the Management Sciences (INFORMS) is a particularly prominent international society in this area. Among its various journals is one called ­Interfaces that regularly publishes articles describing major management science studies and the impact they had on their organizations. Management science (MS) has had numerous applications of various types in business and industry, sometimes resulting in annual savings of millions, or even hundreds of millions, of dollars. As an example, many hundreds of management scientists work on such airline problems as how to most effectively assign airplanes and crews to flights and how to develop fare structures that maximize revenue. For decades, financial services firms have used portfolio selection techniques that were developed by management scientists who won the Nobel Prize in Economics for their work. Management science models have become a core component of the marketing discipline. Multinational corporations rely on MS for guiding the management of their supply chains. There are numerous other examples of MS applications that are having a dramatic impact on the companies involved. Management science also is widely used outside business and industry. For example, it is having an increasing impact in the health care area, with applications involving improved management of health care delivery and operations, disease modeling, clinical diagnosis and decision making, radiation therapy, and so on. Applications of MS also abound at various levels of government, ranging from dealing with national security issues at the federal level to managing the delivery of emergency services at the municipal level. Other key governmental applications involve the use of MS modeling to help guide energy, environmental, and global warming policies. Some of the earliest MS applications were military applications, including logistical planning and war gaming, and these continue today. These are just a sampling of the numerous applications of management science that are having a major impact on the organizations involved. The list goes on and on. To give you a better notion of the wide applicability of management science, we list some actual applications in Table 1.1. Note the diversity of organizations and applications in the first An Application Vignette General Motors (GM) is one of the largest and most successful companies in the world. One major reason for this great success is that GM also is one of the world’s leading users of advanced analytics and management science. In recognition of the great impact that the application of these techniques has had on the success of the company, GM was awarded the 2016 INFORMS Prize. INFORMS (The Institute for Operations Research and the Management Sciences) awards the INFORMS Prize to just one organization each year for its particularly exceptional record in applying advanced analytics and operations research/­ management science (OR/MS) throughout the organization. The award winner must have repeatedly applied these techniques in pioneering, varied, novel, and lasting ways. Following is the citation that describes why GM won the prize for 2016. Citation: The 2016 INFORMS Prize is awarded to General Motors for its sustained record of innovative and impactful applied operations research and advanced analytics. General Motors has hundreds of OR/MS practitioners worldwide who play a vital role in driving data-driven decisions in everything from designing, building, selling, and servicing vehicles to purchasing, logistics and quality. The team is constantly developing new business models and vetting emerging opportunities. GM has developed new market research and analysis techniques to understand what products and features customers most want, to determine the ideal vehicles for their dealers to stock, and to identify the steps they can take to achieve GM’s goal of creating customers for life. GM is also leading the industry by using data science and advanced analytics to predict failure of automotive components and systems before customers are inconvenienced. GM’s ­industry-first Proactive Alert messages notify customers through their OnStar system of a possible malfunction, transforming a potential emergency repair into routine planned maintenance. “Over the last seven decades, OR/MS techniques have been used to improve our understanding of everything from traffic science and supply chain logistics to manufacturing productivity, product development, vehicles telematics and prognostics,” said Gary Smyth, executive director of GM Global R&D Laboratories. “These approaches to problem solving permeate almost everything we do.” The impact OR/MS is now having on its business accelerated in 2007, when GM created a center of expertise for Operations Research to promote best practices and transfer new technologies. It since has expanded to include partner teams in product development, supply chain, finance, information technology and other functions. (Note: The application vignette in Section 11.5 describes just one example of the numerous applications of advanced analytics and management science at GM, where this one led to ­billions of dollars in savings and increased revenue.) “Past INFORMS Awards, 2016: General Motors,” INFORMS.com ©INFORMS. Used with permission. All rights reserved. TABLE 1.1 Applications of Management Science to Be Described in Application Vignettes Organization Area of Application Section Annual Savings General Motors Swift & Company Samsung Electronics Numerous applications Improve sales and manufacturing performance Reduce manufacturing times and inventory levels 1.4 2.1 2.7 INDEVAL Chevron Taylor Communications Welch’s Hewlett-Packard Norwegian companies Canadian Pacific Railway Waste Management MISO Netherlands Railways Continental Airlines Bank Hapoalim Group Settle all securities transactions in Mexico Optimize refinery operations Assign print jobs to printers Optimize use and movement of raw materials Product portfolio management Maximize flow of natural gas through offshore pipeline network Plan routing of rail freight Develop a route-management system for trash collection and disposal Administer the transmission of electricity in 13 states Optimize operation of a railway network Reassign crews to flights when schedule disruptions occur Develop a decision-support system for investment advisors 3.2 3.4 3.6 4.3 6.1 6.3 6.4 7.1 7.2 7.4 7.5 8.2 DHL Workers’ Compensation Board CDC L. L. Bean Taco Bell CSAV General Motors Optimize the use of marketing resources Manage high-risk disability claims and rehabilitation Eradicate polio Forecast staffing needs at call centers Forecast the level of business throughout the day Optimize global shipping Improve the throughput of its production lines 8.4 9.3 9.7 10.2 10.3 10.6 11.5 Not estimated $12 million $200 million more revenue $150 million Nearly $1 billion $10 million $150,000 $180 million $140 million $100 million $100 million $700 million $105 million $40 million $31 million more revenue $260 million $4 million $1.5 billion $300,000 $13 million $81 million $150 million Federal Aviation Administration Sasol Merrill Lynch Manage air traffic flows in severe weather Improve the efficiency of its production processes Pricing analysis for providing financial services 12.2 12.3 13.4 Kroger Pharmacy inventory management 13.8 $200 million $23 million $50 million more revenue $10 million 15 16 Chapter One Introduction two columns. The third column identifies the section where an “application vignette” devotes several paragraphs to describing the application and also references an article that provides full details. (This section includes the first of these application vignettes, where this one mentions a wide variety of important applications.) The last column indicates that these applications typically resulted in annual savings in the many millions of dollars. Furthermore, additional benefits not recorded in the table (e.g., improved service to customers and better managerial control) sometimes were considered to be even more important than these financial benefits. (You will have an opportunity to investigate these less tangible benefits further in Problems 1.9 and 1.10.) A link to the articles in Interfaces that describe these applications in detail is provided at www.mhhe.com/Hillier6e. We are grateful to INFORMS for our special partnership to make these articles available to you through this link. We think you will find these articles interesting and enlightening in illustrating the dramatic impact that management science sometimes can have on the success of a variety of organizations. You also will see a great variety of applications of management science throughout the book in the form of case studies, examples, and end-of-chapter cases. Some of the applications are similar to ones described in application vignettes, but many others are quite different. However, they all generally fall into one of three broad categories, namely, applications in the areas of operations management, finance, and marketing. Tables 1.2, 1.3, and 1.4 list TABLE 1.2 Case Studies and ­Examples in the Area of Operations Management Location Type of Application Sec. 2.1 onward Case 2-1 Case 2-2 Case 2-3 Sec. 3.3 Sec. 3.5 Sec. 3.6 Case 3-1 Case 3-3 Cases 3-5, 5-4, 7-3 Case 3-6 A case study: What is the most profitable mix of products? Which mix of car models should be produced? Which mix of ingredients should go into the casserole in a university cafeteria? Which mix of customer–service agents should be hired to staff a call center? Personnel scheduling of customer–service agents Minimize the cost of shipping a product from factories to customers Optimize the assignment of personnel to tasks How should a product be shipped to market? Which mix of women’s clothing should be produced for next season? Develop a plan for assigning students to schools so as to minimize busing costs Which mixes of solid waste materials should be amalgamated into different grades of a salable product? How should qualified managers be assigned to new R&D projects? Develop and analyze a steel company’s plan for pollution abatement Plan the mix of livestock and crops on a farm with unpredictable weather Minimize the cost of operating a distribution network A case study: Maximize the flow of goods through a distribution network Find the shortest path from an origin to a destination Logistical planning for a military campaign Develop the most profitable flight schedules for an airline Operate and expand a private computer network Choose the best combination of R&D projects to pursue Select the best sites for emergency services facilities Airline crew scheduling Production planning when setup costs are involved Make inventory decisions for a retailer’s warehouse Production planning when overtime is needed Find the shortest route to visit all the American League ballparks Many examples of commercial service systems, internal service systems, and transportation service systems that can be analyzed with queueing models A case study: An analysis of competing proposals for more quickly providing maintenance services to customers Analysis of proposals for reducing in-process inventory Comparison of whether corrective maintenance or preventive maintenance is better A case study: Would it be profitable for the owner of a small business to add an associate? Analysis of proposals for relieving a production bottleneck A case study: How much of a perishable product should be added to a retailer’s inventory? Plan a complex project to ensure a strong likelihood of meeting the project deadline Case 3-7 Case 5-2 Case 5-3 Sec. 6.1 Secs. 6.2, 6.3 Sec. 6.4 Case 6-1 Case 6-3 Cases 6-4 , 7-4 Sec. 7.2 Sec. 7.3 Sec. 7.4 Sec. 7.5 Case 7-2 Sec. 8.3 Sec. 8.5 Sec. 11.2 Sec. 11.4 onward Cases 11-2, 12-2 Sec. 12.1 Secs. 12.2, 12.3 Case 12-1 Sec. 13.1 Sec. 13.3 1.4 The Impact of Management Science 17 TABLE 1.3 Case Studies and Examples in the Area of Finance TABLE 1.4 Case Studies and Examples in the Area of Marketing Location Type of Application Sec. 1.2 Case 1-1 Sec. 3.2 Sec. 3.2 Case 3-2 Sec. 4.1 onward Case 4-1 Sec. 6.4 Case 6-2 Sec. 7.1 Case 7-1 Sec. 8.2 Sec. 8.5 Case 8-2 Case 8-3 Sec. 9.1 onward Case 9-1 Sec. 12.1 Sec. 13.2 Sec. 13.4 Sec. 13.5 Sec. 13.6 Case 13-1 Case 13-2 Break-even analysis Break-even analysis and what-if analysis An airline choosing which airplanes to purchase Capital budgeting of real-estate development projects Develop a schedule for investing in a company’s computer equipment A case study: Develop a financial plan for meeting future cash flow needs Develop an investment and cash flow plan for a pension fund Minimize the cost of car ownership Find the most cost-effective method of converting various foreign currencies into dollars A case study: Determine the most profitable combination of investments Develop an investment plan for purchasing art Portfolio selection that balances expected return and risk Select a portfolio to beat the market as frequently as possible Determine an optimal investment portfolio of stocks Develop a long-range plan to purchase and sell international bonds A case study: Choose whether to drill for oil or sell the land instead Choose a strategy for the game show, “Who Wants to Be a Millionaire?” Analysis of a new gambling game Choose the bid to submit in a competitive bidding process Develop a financial plan when future cash flows are somewhat unpredictable Risk analysis when assessing financial investments How much overbooking should be done in the travel industry? Analysis of how a company’s cash flows might evolve over the next year Calculate the value of a European call option Location Type of Application Secs. 2.7, 3.3 Case 2-1 Secs. 3.1, 3.4 Case 3-4 Case 5-1 Determine the best mix of advertising media Evaluate whether an advertising campaign would be worthwhile A case study: Which advertising plan best achieves managerial goals? Develop a representative marketing survey Analysis of the trade-off between advertising costs and the resulting increase in sales of several products Balance the speed of bringing a new product to market and the associated costs Deal with nonlinear marketing costs Refine the advertising plan developed in the case study presented in Sections 3.1 and 3.4 Should a company immediately launch a new product or test-market it first? Should a company buy additional marketing research before deciding whether to launch a new product? Plan a sequence of decisions for a possible new product A case study: Manage a call center for marketing goods over the telephone Improve forecasts of demand for a call center Estimate customer waiting times for calling into a call center Sec. 6.4 Sec. 8.3 Case 8-1 Case 9-2 Case 9-3 Case 9-4 Sec. 10.2 onward Case 10-1 Case 11-1 these applications in these three respective areas, where the first column identifies where the application is described. In the second column of each table, note the many different ways in which management science can have a real impact in helping improve managerial decisions. Even these long lists of applications in Tables 1.1 to 1.4 are just a sample of the numerous important ways in which management science is applied in organizations around the world. We do not have enough space to provide a more comprehensive compilation of the important applications. (Other applications are included in the supplementary chapters at www.mhhe.com/Hillier6e.) A hallmark of management science is its great flexibility in dealing with new managerial problems as they arise. 18 Chapter One Introduction 1.5 SOME SPECIAL FEATURES OF THIS BOOK The focus of this book is on teaching what an enlightened future manager needs to learn from a management science course. It is not on trying to train technical analysts. This focus has led us to include a number of special features that we hope you enjoy. One special feature is that the entire book revolves around modeling as an aid to managerial decision making. This is what is particularly relevant to a manager. Although they may not use this term, all managers often engage in at least informal modeling (abstracting the essence of a problem to better analyze it), so learning more about the art of modeling is important. Since managers instigate larger management science studies done by others, they also need to be able to recognize the kinds of managerial problems where such a study might be helpful. Thus, a future manager should acquire the ability both to recognize when a management science model might be applicable and to properly interpret the results from analyzing the model. Therefore, rather than spending substantial time in this book on mathematical theory, the mechanics of solution procedures, or the manipulation of spreadsheets, the focus is on the art of model formulation, the role of a model, and the analysis of model results. A wide range of model types is considered. Another special feature is a heavy emphasis on case studies to better convey these ideas in an interesting way in the context of applications. Every subsequent chapter includes at least one case study that introduces and illustrates the application of that chapter’s techniques in a realistic setting. In a few instances, the entire chapter revolves around a case study. Although considerably smaller and simpler than most real studies (to maintain clarity), these case studies are patterned after actual applications requiring a major management science study. Consequently, they convey the whole process of such a study, some of the pitfalls involved, and the complementary roles of the management science team and the manager responsible for the decisions to be made. To complement these case studies, every chapter also includes major cases at the end. These realistic cases can be used for individual assignments, team projects, or case studies in class. In addition, the University of Western Ontario Ivey School of Business (the second largest producer of teaching cases in the world) also has specially selected cases from its case collection that match the chapters in this textbook. These cases are available on the Ivey Website, www.cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. The book also places heavy emphasis on conveying the impressive impact that management science is having on improving the efficiency of numerous organizations around the world. Therefore, you will see many examples of actual applications throughout the book in the form of boxed application vignettes. You then will have the opportunity to learn more about these actual applications by reading the articles fully describing them through using the links provided in www.mhhe.com/Hillier6e. As indicated in Table 1.1, these applications sometimes resulted in annual savings of millions, tens of millions, or even hundreds of millions of dollars. In addition, we try to provide you with a broad perspective about the nature of the real world of management science in practice. It is easy to lose sight of this world when cranking through textbook exercises to master the mechanics of a series of techniques. Therefore, we shift some emphasis from mastering these mechanics to seeing the big picture. The case studies, cases, and descriptions of actual applications are part of this effort. A new feature with this edition is the inclusion of a McGraw-Hill web-based learning tool called Connect. In addition to providing various instructor resources, Connect includes a powerful tool called SmartBook that provides personalized instruction to help students better learn the material. Many students might benefit by using SmartBook. This tool can be accessed at connect.mheducation.com. The preface provides more information about both Connect and SmartBook. Another feature is the inclusion of one or more solved problems for each chapter to help you get started on your homework for that chapter. The statement of each solved problem is given just above the Problems section of the chapter, and then you can find the complete solutions at www.mhhe.com/Hillier6e. As described further in the preface, this same website provides a full range of resources of interest to students, including access to supplementary text material (both supplements to 1.5 Some Special Features of this Book 19 book chapters and additional chapters) and much more. For example, this website includes numerous spreadsheet files for every chapter in this book. Each time a spreadsheet example is presented in the book, a live spreadsheet that shows the formulation and solution for the example also is provided. This gives a convenient reference, or even useful templates, when you set up spreadsheets to solve similar problems. Also, for many models in the book, template spreadsheet files are provided that already include all the equations necessary to solve the model. You simply enter the data for the model and the solution is immediately calculated. The last, but certainly not the least, of the special features of this book is the accompanying software. We will describe and illustrate how to use today’s premier spreadsheet package, Microsoft Excel, to formulate many management science models in a spreadsheet format. Many of the models considered in this book can be solved using standard Excel. Some Excel add-ins also are available to solve other models. Appendix A provides a primer on the use of Excel. Included with standard Excel is an add-in, called Solver, which is used to solve most of the optimization models considered in the first half of this book. Also included with this edition of the textbook is a very powerful software package from Frontline Systems, Inc., called Analytic Solver® for Education (hereafter referred to as Analytic Solver). Instructions for downloading this software are available at www.mhhe.com/Hillier6e. Some special features of Analytic Solver are a significantly enhanced version of the basic Solver included with Excel, the ability to build decision trees within Excel, as covered in Chapter 9, forecasting tools, as covered in Chapter 10, and tools to build computer simulation models within Excel, as covered in Chapter 13. In addition to the Analytic Solver add-in for Excel, you also have access to AnalyticSolver.com. This is cloud-based software accessed through a browser. It has the same look and essentially all the features of Analytic Solver in Excel, but can be used by any computer with access to the Internet without downloading additional software. Most of the software used in this book is compatible with both Excel for Windows PCs and Excel for Macintosh computers (Macs). Analytic Solver is not directly compatible with Macs, although it works well on any recent Mac with Boot Camp or virtualization software. AnalyticSolver.com works through a browser, so can be used with both Windows and Macs (or even other operating systems). For the most up-to-date information on software compatibility and relevant differences between Windows PC versions and Mac versions, please refer to the Software Compatibility link at www.mhhe.com/Hillier6e. We should point out that Excel is not designed for dealing with the really large management science models that occasionally arise in practice. More powerful software packages that are not based on spreadsheets generally are used to solve such models instead. However, management science teams, not managers, primarily use these sophisticated packages (including using modeling languages to help input the large models). Since this book is aimed mainly at future managers rather than future management scientists, we will not have you use these packages. To alert you to relevant learning aids available, the end of each chapter has a list entitled “Learning Aids for This Chapter.” 1.6 Summary Management science is a discipline area that attempts to aid managerial decision making by applying a scientific approach to managerial problems that involve quantitative factors. The rapid development of this discipline began in the 1940s and 1950s. The onslaught of the computer revolution has since continued to give great impetus to its growth. Further impetus now is being provided by the widespread use of spreadsheet software, which greatly facilitates the application of management science by managers and others. A major management science study involves conducting a systematic investigation that includes careful data gathering, developing and testing hypotheses about the problem (typically in the form of a mathematical model), and applying sound logic in the subsequent analysis. The management science team then presents its recommendations to the managers who must make the decisions about how to resolve the problem. Smaller studies might be done by managers themselves with the aid of spreadsheets. A major part of a typical management science study involves incorporating the quantitative factors into a mathematical model (perhaps incorporated into a spreadsheet) and then applying mathematical 20 Chapter One Introduction procedures to solve the model. Such a model uses decision variables to represent the quantifiable decisions to be made. An objective function expresses the appropriate measure of performance in terms of these decision variables. The constraints of the model express the restrictions on the values that can be assigned to the decision variables. The parameters of the model are the constants that appear in the objective function and the constraints. An example involving break-even analysis was used to illustrate a mathematical model. Now that we have entered the era of big data, an exciting development in recent years has been the rise of analytics for transforming data into insight for making better decisions. The increasing momentum of the analytics movement continues to be very impressive. Analytics draws on various quantitative decision sciences. For example, it makes heavy use of data science and forecasting, but the powerful optimization techniques of management science also are particularly key tools. Because management science is at the core of advanced analytics, its usage should continue to grow rapidly. Management science has had an impressive impact on improving the efficiency of numerous organizations around the world. In fact, many award-winning applications have resulted in annual savings in the millions, tens of millions, or even hundreds of millions of dollars. The focus of this book is on emphasizing what an enlightened future manager needs to learn from a management science course. Therefore, the book revolves around modeling as an aid to managerial decision making. Many case studies (within the chapters) and cases (at the end of chapters) are used to better convey these ideas. Glossary break-even point The production and sales volume for a product that must be exceeded to achieve a profit. (Section 1.2), 9 analytics A discipline closely related to ­management science that makes extensive use of data to analyze trends, make forecasts, and apply optimization techniques. (Sections 1.1 and 1.3), 3 and 12-14 Connect A web-based teaching and learning platform that provides both SmartBook and a full range of instructor resources. (Section 1.5), 18 constraint An inequality or equation in a mathematical model that expresses some restrictions on the values that can be assigned to the decision variables. (Section 1.2), 10 decision support system An interactive computerbased system that aids managerial decision making. (Section 1.1), 5 decision variable An algebraic variable that represents a quantifiable decision to be made. (Section 1.2), 8 mathematical model An approximate representation of, for example, a business problem that is expressed in terms of mathematical symbols and expressions. (Section 1.1), 4 Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Excel Files: Special Products Co. Example model An approximate representation of something. (Section 1.1), 4 objective function A mathematical expression in a model that gives the measure of performance for a problem in terms of the decision variables. (Section 1.2), 10 operations research The traditional name for management science that still is widely used outside of business schools. (Section 1.1), 3 parameter One of the constants in a mathematical model. (Section 1.2), 10 range name A descriptive name given to a cell or range of cells that immediately identifies what is there. (Section 1.2), 7 SmartBook A web-based tool within Connect that provides personalized instruction to help students learn the material better. (Section 1.5), 18 spreadsheet model An approximate representation of, for example, a business problem that is laid out on a spreadsheet in a way that facilitates analysis of the problem. (Section 1.1), 4 what-if analysis Analysis of how the recommendations of a model might change if any of the estimates providing the numbers in the model eventually need to be corrected. (Section 1.2), 10 Chapter 1 Solved Problem 21 Solved Problem The solution is available at www.mhhe.com/Hillier6e. 1.S1. Make or Buy? Power Notebooks, Inc., plans to manufacture a new line of notebook computers. Management is trying to decide whether to purchase the LCD screens for the computers from an outside supplier or to manufacture the screens in-house. The screens cost $100 each from the outside supplier. To set up the assembly process required to produce the screens in-house would cost $100,000. The company could then produce each screen for $75. The number of notebooks that eventually will be produced (Q) is unknown at this point. a. Set up a spreadsheet that will display the total cost of both options for any value of Q. Use trial and error with the spreadsheet to determine the range of production volumes for which each alternative is best. b. Use a graphical procedure to determine the break-even point for Q (i.e., the quantity at which both options yield the same cost). c. Use an algebraic procedure to determine the break-even point for Q. Problems 1.1. The manager of a small firm is considering whether to produce a new product that would require leasing some special equipment at a cost of $20,000 per month. In addition to this leasing cost, a production cost of $10 would be incurred for each unit of the product produced. Each unit sold would generate $20 in revenue. Develop a mathematical expression for the monthly profit that would be generated by this product in terms of the number of units produced and sold per month. Then determine how large this number needs to be each month to make it profitable to produce the product. 1.2. Management of the Toys R4U Company needs to decide whether to introduce a certain new novelty toy for the upcoming Christmas season, after which it would be discontinued. The total cost required to produce and market this toy would be $500,000 plus $15 per toy produced. The company would receive revenue of $35 for each toy sold. a. Assuming that every unit of this toy that is produced is sold, write an expression for the profit in terms of the number produced and sold. Then find the breakeven point that this number must exceed to make it worthwhile to introduce this toy. b. Now assume that the number that can be sold might be less than the number produced. Write an expression for the profit in terms of these two numbers. c. Formulate a spreadsheet that will give the profit in part b for any values of the two numbers. d. Write a mathematical expression for the constraint that the number produced should not exceed the number that can be sold. 1.3. A reliable sales forecast has been obtained indicating that the Special Products Company (see Section 1.2) would be able to sell 30,000 iWatches, which appears to be enough to justify introducing this new product. However, management is concerned that this conclusion might change if more accurate estimates were available for the research-and-development cost, the marginal production cost, and the unit revenue. Therefore, before a final decision is made, management wants what-if analysis done on these estimates. Use the spreadsheet from Figure 1.3 (see this chapter’s Excel files at www.mhhe.com/Hillier6e) and trial-and-error to perform what-if analysis by independently investigating each of the following questions. a. How large can the research-and-development cost be before the watches cease to be profitable? b. How large can the marginal production cost be before the watches cease to be profitable? c. How small can the unit revenue be before the watches cease to be profitable? 1.4. Reconsider the problem facing the management of the Special Products Company as presented in Section 1.2. A more detailed investigation now has provided better estimates of the data for the problem. The research-and-development cost still is estimated to be $10 million, but the new estimate of the marginal production cost is $1,300. The revenue from each watch sold now is estimated to be $1,700. a. Use a graphical procedure to find the new breakeven point. b. Use an algebraic procedure to find the new breakeven point. c. State the mathematical model for this problem with the new data. d. Incorporate this mathematical model into a spreadsheet with a sales forecast of 30,000. Use this spreadsheet model to find the new break-even point, and then determine the production quantity and the estimated total profit indicated by the model. e. Suppose that management fears that the sales forecast may be overly optimistic and so does not want to consider producing more than 20,000 watches. Use the spreadsheet from part d to determine what the production quantity should be and the estimated total profit that would result. 1.5. The Best-for-Less Corp. supplies its two retail outlets from its two plants. Plant A will be supplying 30 shipments next month. Plant B has not yet set its production schedule for next month but has the capacity to produce and ship any amount up to a maximum of 50 shipments. Retail outlet 1 has submitted its order for 40 shipments for next month. Retail outlet 2 needs a minimum of 25 shipments next month but would be happy to receive more. The production costs are the same at the two 22 Chapter One Introduction plants but the shipping costs differ. The shipping cost per shipment from each plant to each retail outlet is given below, along with a summary of the other data. The distribution manager, Jennifer Lopez, now needs to develop a plan for how many shipments to send from each plant to each of the retail outlets next month. Her objective is to minimize the total shipping cost. Unit Shipping Cost Plant A Plant B Retail Outlet 1 Retail Outlet 2 Supply $700 $800 $400 $600 = 30 shipments ≤ 50 shipments Needed = 40 shipments ≥ 25 shipments a. Identify the individual decisions that Jennifer needs to make. For each of these decisions, define a decision variable to represent the decision. b. Write a mathematical expression for the total shipping cost in terms of the decision variables. c. Write a mathematical expression for each of the constraints on what the values of the decision variables can be. d. State a complete mathematical model for Jennifer’s problem. e. What do you think Jennifer’s shipping plan should be? Explain your reasoning. Then express your shipping plan in terms of the decision variables. 1.6. The Water Sports Company soon will be producing and marketing a new model line of motor boats. The production manager, Michael Jensen, now is facing a make-or-buy decision regarding the outboard motor to be installed on each of these boats. Based on the total cost involved, should the motors be produced internally or purchased from a vendor? Producing them internally would require an investment of $1 million in new facilities as well as a production cost of $1,600 for each motor produced. If purchased from a vendor instead, the price would be $2,000 per motor. Michael has obtained a preliminary forecast from the company’s marketing division that 3,000 boats in this model line will be sold. a. Use spreadsheets to display and analyze Michael’s two options. Which option should be chosen? b. Michael realizes from past experience that preliminary sales forecasts are quite unreliable, so he wants to check on whether his decision might change if a more careful forecast differed significantly from the preliminary forecast. Determine a break-even point for the production and sales volume below which the buy option is better and above which the make option is better. 1.7. Reconsider the Special Products Company problem presented in Section 1.2. Although the company is well qualified to do most of the work in producing the iWatch, it currently lacks much expertise in one key area, namely, developing and producing a miniature camera to be embedded into the iWatch. Therefore, management now is considering contracting out this part of the job to another company that has this expertise. If this were done, the Special Products Company would reduce its research-and-development cost to $5 million, as well as reduce its marginal production cost to $750. However, the Special Products Company also would pay this other company $500 for each miniature camera and so would incur a total marginal cost of $1,250 (including its payment to the other company) while still obtaining revenue of $2,000 for each watch produced and sold. However, if the company does all the production itself, all the data presented in Section 1.2 still apply. After obtaining an analysis of the sales potential, management believes that 30,000 watches can be sold. Management now wants to determine whether the make option (do all the development and production internally) or the buy option (contract out the development and production of the miniature cameras) is better. a. Use a spreadsheet to display and analyze the buy option. Show the relevant data and financial output, including the total profit that would be obtained by producing and selling 30,000 watches. b. Figure 1.3 shows the analysis for the make option. Compare these results with those from part a to determine which option (make or buy) appears to be better. c. Another way to compare these two options is to find a break-even point for the production and sales volume, below which the buy option is better and above which the make option is better. Begin this process by developing an expression for the difference in profit between the make and buy options in terms of the number of grandfather clocks to produce for sale. Thus, this expression should give the incremental profit from choosing the make option rather than the buy option, where this incremental profit is 0 if 0 watches are produced but otherwise is negative below the break-even point and positive above the break-even point. Using this expression as the objective function, state the overall mathematical model (including constraints) for the problem of determining whether to choose the make option and, if so, how many units of the LCD display (one per watch) to produce. d. Use a graphical procedure to find the break-even point described in part c. e. Use an algebraic procedure to find the break-even point described in part c. f. Use a spreadsheet model to find the break-even point described in part c. What is the conclusion about what the company should do? 1.8. Select one of the applications of management science listed in Table 1.1. Read the article that is referenced in the application vignette presented in the section shown in the third column. (A link to all these articles is at www.mhhe.com/­Hillier6e.) Write a two-page summary of the application and the benefits (including nonfinancial benefits) it provided. 1.9. Select three of the applications of management science listed in Table 1.1. For each one, read the article that is referenced in the application vignette presented in the section shown in the third column. (A link to all these articles is at www.mhhe.com/Hillier6e.) For each one, write a one-page summary of the application and the benefits (including nonfinancial benefits) it provided. Case 1-1 Keeping Time 23 Case 1-1 Keeping Time Founded nearly 50 years ago by Alfred Lester-Smith, ­Beautiful Clocks specializes in developing and marketing a diverse line of large ornamental clocks for the finest homes. Tastes have changed over the years, but the company has prospered by continually updating its product line to satisfy its affluent clientele. The Lester-Smith family continues to own a majority share of the company and the grandchildren of Alfred Lester-Smith now hold several of the top managerial positions. One of these grandchildren is Meredith Lester-Smith, the new CEO of the company. Meredith feels a great responsibility to maintain the family heritage with the company. She realizes that the company needs to continue to develop and market exciting new products. Since the 50th anniversary of the founding of the company is rapidly approaching, she has decided to select a particularly special new product to launch with great fanfare on this anniversary. But what should it be? As she ponders this crucial decision, Meredith’s thoughts go back to the magnificent grandfather clock that her grandparents had in their home many years ago. She had admired the majesty of that clock as a child. How about launching a modern version of this clock? This is a difficult decision. Meredith realizes that grandfather clocks now are largely out of style. However, if she is so nostalgic about the memory of the grandfather clock in her grandparents’ home, wouldn’t there be a considerable number of other relatively wealthy couples with similar memories who would welcome the prestige of adding the grandeur of a beautifully designed limited-edition grandfather clock in their home? Maybe. This also would highlight the heritage and continuity of the company. It all depends on whether there would be enough sales potential to make this a profitable product. Meredith had an excellent management science course as part of her MBA program in college, so she realizes that break-even analysis is needed to help make this decision. With this in mind, she instructs several staff members to investigate this prospective product further, including developing estimates of the related costs and revenues as well as forecasting the potential sales. One month later, the preliminary estimates of the relevant financial figures come back. The cost of designing the grandfather clock and then setting up the production facilities to produce this product would be approximately $250,000. There would be only one production run for this limited-edition grandfather clock. The additional cost for each clock produced would be roughly $2,000. The marketing department estimates that their price for selling the clocks can be successfully set at about $4,500 apiece, but a firm forecast of how many clocks can be sold at this price has not yet been obtained. However, it is believed that the sales likely would reach into three digits. Meredith wants all these numbers pinned down considerably further. However, she feels that some analysis can be done now to draw preliminary conclusions. a. Assuming that all clocks produced are sold, develop a spreadsheet model for estimating the profit or loss from producing any particular number of clocks. b. Use this spreadsheet to find the break-even point by trial and error. c. Develop the corresponding mathematical expression for the estimated profit in terms of the number of clocks produced. d. Use a graphical procedure to find the break-even point. e. Use the algebraic procedure to find the break-even point. A fairly reliable forecast now has been obtained indicating that the company would be able to sell 300 of the limited-­edition grandfather clocks, which appears to be enough to justify introducing this new product. However, Meredith is concerned that this conclusion might change if more accurate estimates were available for the various costs and revenues. Therefore, she wants what-if analysis done on these estimates. Perform whatif analysis by independently investigating each of the following questions. f. How large can the cost of designing this product and setting up the production facilities be before the grandfather clocks cease to be profitable? g. How large can the production cost for each additional clock be before the grandfather clocks cease to be profitable? h. If both of the costs identified in parts f and g were 50% larger than their initial estimates, would producing and selling the grandfather clocks still be profitable? i. How small can the price for selling each clock be before the grandfather clocks cease to be profitable? Now suppose that 300 grandfather clocks are produced but only 200 are sold. j. Would it still be profitable to produce and sell the grandfather clocks under this circumstance? Additional Case An additional case for this chapter is also available at the University of Western Ontario Ivey School of Business Website, cases. ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. Chapter Two Linear Programming: Basic Concepts Learning Objectives After completing this chapter, you should be able to 1. Explain what linear programming is. 2. Identify the three key questions to be addressed in formulating any spreadsheet model. 3. Name and identify the purpose of the four kinds of cells used in linear programming spreadsheet models. 4. Formulate a basic linear programming model in a spreadsheet from a description of the problem. 5. Present the algebraic form of a linear programming model from its formulation on a spreadsheet. 6. Apply the graphical method to solve a two-variable linear programming problem. 7. Use Excel to solve a linear programming spreadsheet model. The management of any organization regularly must make decisions about how to allocate its resources to various activities to best meet organizational objectives. Linear programming is a powerful problem-solving tool that aids management in making such decisions. It is applicable to both profit-making and not-for-profit organizations, as well as governmental agencies. The resources being allocated to activities can be, for example, money, different kinds of personnel, and different kinds of machinery and equipment. In many cases, a wide variety of resources must be allocated simultaneously. The activities needing these resources might be various production activities (e.g., producing different products), marketing activities (e.g., advertising in different media), financial activities (e.g., making capital investments), or some other activities. Some problems might even involve activities of all these types (and perhaps others), because they are competing for the same resources. You will see as we progress that even this description of the scope of linear programming is not sufficiently broad. Some of its applications go beyond the allocation of resources. However, activities always are involved. Thus, a recurring theme in linear programming is the need to find the best mix of activities—which ones to pursue and at what levels. Like the other management science techniques, linear programming uses a mathematical model to represent the problem being studied. The word linear in the name refers to the form of the mathematical expressions in this model. Programming does not refer to computer programming; rather, it is essentially a synonym for planning. Thus, linear programming means the planning of activities represented by a linear mathematical model. Because it comprises a major part of management science, linear programming takes up several chapters of this book. Furthermore, many of the lessons learned about how to apply linear programming also will carry over to the application of other management science techniques. This chapter focuses on the basic concepts of linear programming. 24 2.1 A Case Study: The Wyndor Glass Co. Product-Mix Problem 25 2.1 A CASE STUDY: THE WYNDOR GLASS CO. PRODUCT-MIX PROBLEM Jim Baker has had an excellent track record during his seven years as manager of new product development for the Wyndor Glass Company. Although the company is a small one, it has been experiencing considerable growth largely because of the innovative new products developed by Jim’s group. Wyndor’s president, John Hill, has often acknowledged publicly the key role that Jim has played in the recent success of the company. Therefore, John felt considerable confidence six months ago in asking Jim’s group to develop the following new products: • An 8-foot glass door with aluminum framing. • A 4-foot × 6-foot double-hung, wood-framed window. Although several other companies already had products meeting these specifications, John felt that Jim would be able to work his usual magic in introducing exciting new features that would establish new industry standards. Background The Wyndor Glass Co. produces high-quality glass products, including windows and glass doors that feature handcrafting and the finest workmanship. Although the products are expensive, they fill a market niche by providing the highest quality available in the industry for the most discriminating buyers. The company has three plants that simultaneously produce the components of its products. Plant 1 produces aluminum frames and hardware. Plant 2 produces wood frames. Plant 3 produces the glass and assembles the windows and doors. Because of declining sales for certain products, top management has decided to revamp the company’s product line. Unprofitable products are being discontinued, releasing production capacity to launch the two new products developed by Jim Baker’s group if management approves their release. The 8-foot glass door requires some of the production capacity in Plants 1 and 3, but not Plant 2. The 4-foot × 6-foot double-hung window needs only Plants 2 and 3. Management now needs to address two issues: 1. Should the company go ahead with launching these two new products? 2. If so, what should be the product mix—the number of units of each produced per week— for the two new products? Management’s Discussion of the Issues Having received Jim Baker’s memorandum describing the two new products, John Hill now has called a meeting to discuss the current issues. In addition to John and Jim, the meeting includes Bill Tasto, vice president for manufacturing, and Ann Lester, vice president for marketing. Let’s eavesdrop on the meeting. John Hill (president): Bill, we will want to rev up to start production of these products as soon as we can. About how much production output do you think we can achieve? Bill Tasto (vice president for manufacturing): We do have a little available production capacity, because of the products we are discontinuing, but not a lot. We should be able to achieve a production rate of a few units per week for each of these two products. John: Is that all? Bill: Yes. These are complicated products requiring careful crafting. And, as I said, we don’t have much production capacity available. John: Ann, will we be able to sell several of each per week? Ann Lester (vice president for marketing): Easily. An Application Vignette Swift & Company is a diversified protein-producing business based in Greeley, Colorado. With annual sales of over $8 billion, beef and related products are by far the largest portion of the company’s business. To improve the company’s sales and manufacturing performance, upper management concluded that it needed to achieve three major objectives. One was to enable the company’s customer service representatives to talk to their more than 8,000 customers with accurate information about the availability of current and future inventory while considering requested delivery dates and maximum product age upon delivery. A second was to produce an efficient shift-level schedule for each plant over a 28-day horizon. A third was to accurately determine whether a plant can ship a requested order-line-item quantity on the requested date and time The issue is to find the most profitable mix of the two new products. given the availability of cattle and constraints on the plant’s capacity. To meet these three challenges, a management science team developed an integrated system of 45 linear programming models based on three model formulations to dynamically schedule its beef-fabrication operations at five plants in real time as it receives orders. The total audited benefits realized in the first year of operation of this system were $12.74 million, including $12 million due to optimizing the product mix. Other benefits include a reduction in orders lost, a reduction in price discounting, and better on-time delivery. Source: A. Bixby, B. Downs, and M. Self, “A Scheduling and Capableto-Promise Application for Swift & Company,” Interfaces 36, no. 1 (January–February 2006), pp. 69–86. (A link to this article is available at www.mhhe.com/Hillier6e.) John: Good. Now there’s one more issue to resolve. With this limited production capacity, we need to decide how to split it between the two products. Do we want to produce the same number of both products? Or mostly one of them? Or even just produce as much as we can of one and postpone launching the other one for a little while? Jim Baker (manager of new product development): It would be dangerous to hold one of the products back and give our competition a chance to scoop us. Ann: I agree. Furthermore, launching them together has some advantages from a marketing standpoint. Since they share a lot of the same special features, we can combine the advertising for the two products. This is going to make a big splash. John: OK. But which mixture of the two products is going to be most profitable for the company? Bill: I have a suggestion. John: What’s that? Bill: A couple times in the past, our Management Science Group has helped us with these same kinds of product-mix decisions, and they’ve done a good job. They ferret out all the relevant data and then dig into some detailed analysis of the issue. I’ve found their input very helpful. And this is right down their alley. John: Yes, you’re right. That’s a good idea. Let’s get our Management Science Group working on this issue. Bill, will you coordinate with them? The meeting ends. The Management Science Group Begins Its Work At the outset, the Management Science Group spends considerable time with Bill Tasto to clarify the general problem and specific issues that management wants addressed. A particular concern is to ascertain the appropriate objective for the problem from management’s viewpoint. Bill points out that John Hill posed the issue as determining which mixture of the two products is going to be most profitable for the company. Therefore, with Bill’s concurrence, the group defines the key issue to be addressed as follows. Question: Which combination of production rates (the number of units produced per week) for the two new products would maximize the total profit from both of them? 26 The group also concludes that it should consider all possible combinations of production rates of both new products permitted by the available production capacities in the three plants. For example, one alternative (despite Jim Baker’s and Ann Lester’s objections) is to forgo producing one of the products for now (thereby setting its production rate equal to zero) in order to produce as much as possible of the other product. (We must not neglect the possibility that maximum profit from both products might be attained by producing none of one and as much as possible of the other.) 2.2 Formulating the Wyndor Problem on a Spreadsheet 27 TABLE 2.1 Production Time Used for Each Unit Produced Data for the Wyndor Glass Co. Product-Mix Problem Plant Doors Windows Available per Week 1 2 3 1 hour 0 3 hours 0 2 hours 2 hours 4 hours 12 hours 18 hours Unit profit $300 $500 The Management Science Group next identifies the information it needs to gather to conduct this study: 1. Available production capacity in each of the plants. 2. How much of the production capacity in each plant would be needed by each product. 3. Profitability of each product. Concrete data are not available for any of these quantities, so estimates have to be made. Estimating these quantities requires enlisting the help of key personnel in other units of the company. Bill Tasto’s staff develops the estimates that involve production capacities. Specifically, the staff estimates that the production facilities in Plant 1 needed for the new kind of doors will be available approximately four hours per week. (During the rest of the time, Plant 1 will continue with current products.) The production facilities in Plant 2 will be available for the new kind of windows about 12 hours per week. The facilities needed for both products in Plant 3 will be available approximately 18 hours per week. The amount of each plant’s production capacity actually used by each product depends on its production rate. It is estimated that each door will require one hour of production time in Plant 1 and three hours in Plant 3. For each window, about two hours will be needed in Plant 2 and two hours in Plant 3. By analyzing the cost data and the pricing decision, the Accounting Department estimates the profit from the two products. The projection is that the profit per unit will be $300 for the doors and $500 for the windows. Table 2.1 summarizes the data now gathered. The Management Science Group recognizes this as being a classic product-mix ­problem. Therefore, the next step is to develop a mathematical model—that is, a linear programming model—to represent the problem so that it can be solved mathematically. The next four sections focus on how to develop this model and then how to solve it to find the most profitable mix between the two products, assuming the estimates in Table 2.1 are accurate. Review Questions 2.2 1. 2. 3. 4. What were the two issues addressed by management? The Management Science Group was asked to help analyze which of these issues? How did this group define the key issue to be addressed? What information did the group need to gather to conduct its study? FORMULATING THE WYNDOR PROBLEM ON A SPREADSHEET Spreadsheets provide a powerful and intuitive tool for displaying and analyzing many management problems. We now will focus on how to do this for the Wyndor problem with the popular spreadsheet package Microsoft Excel.1 1 Other spreadsheet packages with similar capabilities also are available, and the basic ideas presented here are still applicable. 28 Chapter Two Linear Programming: Basic Concepts FIGURE 2.1 The initial spreadsheet for the Wyndor problem after transferring the data in Table 2.1 into data cells. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 Hours 6 Hours Used per Unit Produced Available 7 Plant 1 1 0 4 8 Plant 2 0 2 12 9 Plant 3 3 2 18 Formulating a Spreadsheet Model for the Wyndor Problem Excel Tip: Cell shading and borders can be added by using the borders button and the fill color button in the Font Group of the Home tab. Excel Tip: See the margin notes in Section 1.2 for tips on adding range names. These are the three key questions to be addressed in formulating any spreadsheet model. Figure 2.1 displays the Wyndor problem by transferring the data in Table 2.1 onto a spreadsheet. (Columns E and F are being reserved for later entries described below.) We will refer to the cells showing the data as data cells. To distinguish the data cells from other cells in the spreadsheet, they are shaded light blue. (In the textbook figures, the light blue shading appears as light gray.) The spreadsheet is made easier to interpret by using range names. (As mentioned in Section 1.2, a range name is simply a descriptive name given to a cell or range of cells that immediately identifies what is there. Excel allows you to use range names instead of the corresponding cell addresses in Excel equations, since this usually makes the equations much easier to interpret at a glance.) The data cells in the Wyndor Glass Co. problem are given the range names UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and HoursAvailable (G7:G9). To enter a range name, first select the range of cells, then click in the name box on the left of the formula bar above the spreadsheet and type a name. (See ­Appendix A for further details about defining and using range names.) Three questions need to be answered to begin the process of using the spreadsheet to formulate a mathematical model (in this case, a linear programming model) for the problem. 1. What are the decisions to be made? 2. What are the constraints on these decisions? 3. What is the overall measure of performance for these decisions? The preceding section described how Wyndor’s Management Science Group spent considerable time with Bill Tasto, vice president for manufacturing, to clarify management’s view of their problem. These discussions provided the following answers to these questions. Some students find it helpful to organize their thoughts by answering these three key questions before beginning to formulate the spreadsheet model. The changing cells contain the decisions to be made. 1. The decisions to be made are the production rates (number of units produced per week) for the two new products. 2. The constraints on these decisions are that the number of hours of production time used per week by the two products in the respective plants cannot exceed the number of hours available. 3. The overall measure of performance for these decisions is the total profit per week from the two products. Figure 2.2 shows how these answers can be incorporated into the spreadsheet. Based on the first answer, the production rates of the two products are placed in cells C12 and D12 to locate them in the columns for these products just under the data cells. Since we don’t know yet what these production rates should be, they are just entered as zeroes in Figure 2.2. (Actually, any trial solution can be entered, although negative production rates should be excluded since they are impossible.) Later, these numbers will be changed while seeking the best mix of production rates. Therefore, these cells containing the decisions to be made are called changing cells. To highlight the changing cells, they are shaded bright yellow with a light border. (In the textbook figures, the bright yellow appears as gray.) The changing cells are given the range name UnitsProduced (C12:D12). 2.2 Formulating the Wyndor Problem on a Spreadsheet 29 FIGURE 2.2 The complete spreadsheet for the Wyndor problem with an initial trial solution (both production rates equal to zero) entered into the changing cells (C12 and D12). A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0 ≤ 4 8 Plant 2 0 2 0 ≤ 12 9 Plant 3 3 2 0 ≤ 18 Doors Windows Total Profit 0 0 $0 10 11 12 The colon in C7:D9 is Excel shorthand for the range from C7 to D9; that is, the entire block of cells in column C or D and in row 7, 8, or 9. Units Produced Using the answer to the second question, the total number of hours of production time used per week by the two products in the respective plants is entered in cells E7, E8, and E9, just to the right of the corresponding data cells. The total number of production hours depends on the production rates of the two products, so this total is zero when the production rates are zero. With positive production rates, the total number of production hours used per week in a plant is the sum of the production hours used per week by the respective products. The production hours used by a product is the number of hours needed for each unit of the product times the number of units being produced. Therefore, when positive numbers are entered in cells C12 and D12 for the number of doors and windows to produce per week, the data in cells C7:D9 are used to calculate the total production hours per week as follows: Production hours in Plant 1 = 1(# of doors) + 0(# of windows) Production hours in Plant 2 = 0(# of doors) + 2(# of windows) Production hours in Plant 3 = 3(# of doors) + 2(# of windows) Consequently, the Excel equations for the three cells in column E are E7 = C7*C12 + D7*D12 E8 = C8*C12 + D8*D12 E9 = C9*C12 + D9*D12 Output cells show quantities that are calculated from the changing cells. The SUMPRODUCT function is used extensively in linear programming spreadsheet models. where each asterisk denotes multiplication. Since each of these cells provides output that depends on the changing cells (C12 and D12), they are called output cells. Notice that each of the equations for the output cells involves the sum of two products. There is a function in Excel called SUMPRODUCT that will sum up the product of each of the individual terms in two different ranges of cells when the two ranges have the same number of rows and the same number of columns. Each product being summed is the product of a term in the first range and the term in the corresponding location in the second range. For example, consider the two ranges, C7:D7 and C12:D12, so that each range has one row and two columns. In this case, SUMPRODUCT (C7:D7, C12:D12) takes each of the individual terms in the range C7:D7, multiplies them by the corresponding term in the range C12:D12, and then sums up these individual products, just as shown in the first equation above. Applying the range name for UnitsProduced (C12:D12), the formula for E7 becomes =SUMPRODUCT(C7:D7, UnitsProduced). Although optional with such short equations, the SUMPRODUCT function is especially handy as a shortcut for entering longer equations. The formulas in the output cells E7:E9 are very similar. Rather than typing each of these formulas separately into the three cells, it is quicker (and less prone to typos) to type the formula just once in E7 and then copy the formula down into cells E8 and E9. To do this, first enter the formula = SUMPRODUCT(C7:D7, UnitsProduced) in cell E7. Then select cell 30 Chapter Two Linear Programming: Basic Concepts You can make the column absolute and the row relative (or vice versa) by putting a $ sign in front of only the letter (or number) of the cell reference. One easy way to enter a ≤ (or ≥) in a spreadsheet is to type < (or >) with underlining turned on. Excel Tip: After entering a cell reference, repeatedly pressing the F4 key (or command-T on a Mac) will rotate among the four possibilities of relative and absolute references (e.g., C12, $C$12, C$12, $C12). E7 and drag the fill handle (the small box on the lower right corner of the cell cursor) down through cells E8 and E9. When copying formulas, it is important to understand the difference between relative and absolute references. In the formula in cell E7, the reference to cells C7:D7 is based upon the relative position to the cell containing the formula. In this case, the relative position is the two cells in the same row and immediately to the left. This is known as a relative reference. When this formula is copied to new cells using the fill handle, the reference is automatically adjusted to refer to the new cell(s) at the same relative location (the two cells in the same row and immediately to the left). The formula in E8 becomes = SUMPRODUCT(C8:D8, UnitsProduced) and the formula in E9 becomes = SUMPRODUCT(C9:D9, UnitsProduced). This is exactly what we want, since we always want the hours used at a given plant to be based upon the hours used per unit produced at that same plant (the two cells in the same row and immediately to the left). In contrast, the reference to the UnitsProduced in E7 is called an absolute reference. These references do not change when they are filled into other cells but instead always refer to the same absolute cell locations. To make a relative reference, simply enter the cell addresses (e.g., C7:D7). By contrast, references referred to by a range name are treated as absolute references. Another way to make an absolute reference to a range of cells is to put $ signs in front of the letter and number of the cell reference (e.g., $C$12:$D$12). See Appendix A for more details about relative and absolute referencing and copying formulas. Next, ≤ signs are entered in cells F7, F8, and F9 to indicate that each total value to their left cannot be allowed to exceed the corresponding number in column G. (On the computer ≤ (or ≥) is often represented as < = (or > =), since there is no ≤ (or ≥) key on the keyboard.) The spreadsheet still will allow you to enter trial solutions that violate the ≤ signs. However, these ≤ signs serve as a reminder that such trial solutions need to be rejected if no changes are made in the numbers in column G. Finally, since the answer to the third question is that the overall measure of performance is the total profit from the two products, this profit (per week) is entered in cell G12. Much like the numbers in column E, it is the sum of products. Since cells C4 and D4 give the profit from each door and window produced, the total profit per week from these products is Profit = $300(# of doors) + $500(# of windows) Hence, the equation for cell G12 is G12 = SUMPRODUCT(C4:D4, C12:D12) Utilizing range names of TotalProfit (G12), UnitProfit (C4:D4), and UnitsProduced (C12:D12), this equation becomes TotalProfit = SUMPRODUCT(UnitProfit, UnitsProduced) The objective cell contains the overall measure of performance for the decisions in the changing cells. This is a good example of the benefit of using range names for making the resulting equation easier to interpret. TotalProfit (G12) is a special kind of output cell. It is our objective to make this cell as large as possible when making decisions regarding production rates. Therefore, TotalProfit (G12) is referred to as the objective cell. This cell is shaded orange with a heavy border. (In the textbook figures, the orange appears as gray and is distinguished from the changing cells by its darker shading and heavy border.) The bottom of Figure 2.3 summarizes all the formulas that need to be entered in the Hours Used column and in the Total Profit cell. Also shown is a summary of the range names (in alphabetical order) and the corresponding cell addresses. This completes the formulation of the spreadsheet model for the Wyndor problem. With this formulation, it becomes easy to analyze any trial solution for the production rates. Each time production rates are entered in cells C12 and D12, Excel immediately calculates the output cells for hours used and total profit. For example, Figure 2.4 shows the spreadsheet when the production rates are set at four doors per week and three windows per week. Cell G12 shows that this yields a total profit of $2,700 per week. Also note that E7 = G7, E8 < G8, and E9 = G9, so the ≤ signs in column F are all satisfied. Thus, this trial 2.2 Formulating the Wyndor Problem on a Spreadsheet 31 FIGURE 2.3 The spreadsheet model for the Wyndor problem, including the formulas for the objective cell TotalProfit (G12) and the other output cells in column E, where the goal is to maximize the objective cell. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 Unit Profit 4 Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0 ≤ 4 8 Plant 2 0 2 0 ≤ 12 9 Plant 3 3 2 0 ≤ 18 Doors Windows Total Profit 0 0 $0 10 11 12 Units Produced Range Name Cell HoursAvailable HoursUsed HoursUsedPerUnitProduced TotalProfit UnitProfit UnitsProduced G7:G9 E7:E9 C7:D9 G12 C4:D4 C12:D12 E 5 Hours 6 7 8 9 Used =SUMPRODUCT(C7:D7, UnitsProduced) =SUMPRODUCT(C8:D8, UnitsProduced) =SUMPRODUCT(C9:D9, UnitsProduced) G 11 12 FIGURE 2.4 The spreadsheet for the Wyndor problem with a new trial solution entered into the changing cells, UnitsProduced (C12:D12). A 1 B C Total Profit =SUMPRODUCT(UnitProfit, UnitsProduced) D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 8 Plant 2 0 9 Plant 3 3 Doors Windows Total Profit 4 3 $2,700 4 ≤ 4 2 6 ≤ 12 2 18 ≤ 18 10 11 12 Units Produced solution is feasible. However, it would not be feasible to further increase both production rates, since this would cause E7 > G7 and E9 > G9. Does this trial solution provide the best mix of production rates? Not necessarily. It might be possible to further increase the total profit by simultaneously increasing one production rate and decreasing the other. However, it is not necessary to continue using trial and error to explore such possibilities. We shall describe in Section 2.5 how the Excel Solver can be used to quickly find the best (optimal) solution. This Spreadsheet Model Is a Linear Programming Model The spreadsheet model displayed in Figure 2.3 is an example of a linear programming model. The reason is that it possesses all the following characteristics. 32 Chapter Two Linear Programming: Basic Concepts Characteristics of a Linear Programming Model on a Spreadsheet 1. Decisions need to be made on the levels of a number of activities, so changing cells are used to display these levels. (The two activities for the Wyndor problem are the production of the two new products, so the changing cells display the number of units produced per week for each of these products.) 2. These activity levels can have any value (including fractional values) that satisfy a number of constraints. (The production rates for Wyndor’s new products are restricted only by the constraints on the number of hours of production time available in the three plants.) 3. Each constraint describes a restriction on the feasible values for the levels of the activities, where a constraint commonly is displayed by having an output cell on the left, a mathematical sign (≤, ≥, or =) in the middle, and a data cell on the right. (Wyndor’s three constraints involving hours available in the plants are displayed in Figures 2.2–2.4 by having output cells in column E, ≤ signs in column F, and data cells in column G.) 4. The decisions on activity levels are to be based on an overall measure of performance, which is entered in the objective cell. The goal is to either maximize the objective cell or minimize the objective cell, depending on the nature of the measure of performance. (Wyndor’s overall measure of performance is the total profit per week from the two new products, so this measure has been entered in the objective cell G12, where the goal is to maximize this objective cell.) 5. The Excel equation for each output cell (including the objective cell) can be expressed as a SUMPRODUCT function,2 where each term in the sum is the product of a data cell and a changing cell. (The bottom of Figure 2.3 shows how a SUMPRODUCT function is used for each output cell for the Wyndor problem.) Linear programming models are not the only models that can have characteristics 1, 3, and 4. However, characteristics 2 and 5 are key assumptions of linear programming. Therefore, these are the two key characteristics that together differentiate a linear programming model from other kinds of mathematical models that can be formulated on a spreadsheet. Characteristic 2 rules out situations where the activity levels need to have integer values. For example, such a situation would arise in the Wyndor problem if the decisions to be made were the total numbers of doors and windows to produce (which must be integers) rather than the numbers per week (which can have fractional values since a door or window can be started in one week and completed in the next week). When the activity levels do need to have integer values, a similar kind of model (called an integer programming model) is used instead by making a small adjustment on the spreadsheet, as will be illustrated in Section 3.2. Characteristic 5 describes the so-called proportionality assumption of linear programming, which states that each term in an output cell must be proportional to a particular changing cell. This prohibits those cases where the Excel equation for an output cell cannot be expressed as a SUMPRODUCT function. To illustrate such a case, suppose that the weekly profit from producing Wyndor’s new windows can be more than doubled by doubling the production rate because of economies in marketing larger amounts. This would mean that this weekly profit is not simply proportional to this production rate, so the Excel equation for the objective cell would need to be more complicated than a SUMPRODUCT function. Consideration of how to formulate such models will be deferred to Chapter 8. Additional examples of formulating linear programming models on a spreadsheet are provided in Section 2.7 and in both of this chapter’s solved problems presented at the end of the chapter. (All of these examples also will illustrate the procedure described in the next section for using algebra to formulate linear programming models.) Summary of the Formulation Procedure The procedure used to formulate a linear programming model on a spreadsheet for the Wyndor problem can be adapted to many other problems as well. Here is a summary of the steps involved in the procedure. 1. Gather the data for the problem (such as summarized in Table 2.1 for the Wyndor problem). 2. Enter the data into data cells on a spreadsheet. 2 There also are some special situations where a SUM function can be used instead because all the ­numbers that would have gone into the corresponding data cells are 1’s. 2.3 The Mathematical Model in the Spreadsheet 33 3. Identify the decisions to be made on the levels of activities and designate changing cells for displaying these decisions. 4. Identify the constraints on these decisions and introduce output cells as needed to specify these constraints. 5. Choose the overall measure of performance to be entered into the objective cell. 6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (including the objective cell). This procedure does not spell out the details of how to set up the spreadsheet. There generally are alternative ways of doing this rather than a single “right” way. One of the great strengths of spreadsheets is their flexibility for dealing with a wide variety of problems. Review Questions 2.3 1. What are the three questions that need to be answered to begin the process of formulating a linear programming model on a spreadsheet? 2. What are the roles for the data cells, the changing cells, the output cells, and the objective cell when formulating such a model? 3. What is the form of the Excel equation for each output cell (including the objective cell) when formulating such a model? THE MATHEMATICAL MODEL IN THE SPREADSHEET A linear programming model can be formulated either as a spreadsheet model or as an algebraic model. There are two widely used methods for formulating a linear programming model. One is to formulate it directly on a spreadsheet, as described in the preceding section. The other is to use algebra to present the model. The two versions of the model are equivalent. The only difference is whether the language of spreadsheets or the language of algebra is used to describe the model. Both versions have their advantages, and it can be helpful to be bilingual. For example, the two versions lead to different, but complementary, ways of analyzing problems like the Wyndor problem (as discussed in the next two sections). Since this book emphasizes the spreadsheet approach, we will only briefly describe the algebraic approach. Formulating the Wyndor Model Algebraically The reasoning for the algebraic approach is similar to that for the spreadsheet approach. In fact, except for making entries on a spreadsheet, the initial steps are just as described in the preceding section for the Wyndor problem. 1. Gather the relevant data (Table 2.1 in Section 2.1). 2. Identify the decisions to be made (the production rates for the two new products). 3. Identify the constraints on these decisions (the production time used in the respective plants cannot exceed the amount available). 4. Identify the overall measure of performance for these decisions (the total profit from the two products). 5. Convert the verbal description of the constraints and measure of performance into quantitative expressions in terms of the data and decisions (see below). To start performing step 5, note that Table 2.1 indicates that the number of hours of production time available per week for the two new products in the respective plants are 4, 12, and 18. Using the data in this table for the number of hours used per door or window produced then leads to the following quantitative expressions for the constraints: Plant 1: (# of doors) ≤ 4 Plant 2: 2(# of windows) ≤ 12 Plant 3: 3(# of doors) + 2(# of windows) ≤ 18 In addition, negative production rates are impossible, so two other constraints on the decisions are (# of doors) ≥ 0 (# of windows) ≥ 0 The overall measure of performance has been identified as the total profit from the two products. Since Table 2.1 gives the unit profits for doors and windows as $300 and $500, 34 Chapter Two Linear Programming: Basic Concepts respectively, the expression obtained in the preceding section for the total profit per week from these products is Profit = $300(# of doors) + $500(# of windows) The goal is to make the decisions (number of doors and number of windows) so as to maximize this profit, subject to satisfying all the constraints identified above. To state this objective in a compact algebraic model, we introduce algebraic symbols to represent the measure of performance and the decisions. Let P = Profit (total profit per week from the two products, in dollars) D = # of doors (number of the special new doors to be produced per week) W = # of windows (number of the special new windows to be produced per week) Substituting these symbols into the above expressions for the constraints and the measure of performance (and dropping the dollar signs in the latter expression), the linear programming model for the Wyndor problem now can be written in algebraic form as shown below. Algebraic Model Choose the values of D and W so as to maximize P = 300D + 500W subject to satisfying all the following constraints: D ≤ 4 2W ≤ 12 3D + 2W ≤ 18 and D ≥ 0 W ≥ 0 Terminology for Linear Programming Models Much of the terminology of algebraic models also is sometimes used with spreadsheet models. Here are the key terms for both kinds of models in the context of the Wyndor problem. 1. D and W (or C12 and D12 in Figure 2.3) are the decision variables. 2. 300D + 500W [or SUMPRODUCT (UnitProfit, UnitsProduced)] is the objective function. 3. P (or G12) is the value of the objective function (or objective value for short). 4. D ≥ 0 and W ≥ 0 (or C12 ≥ 0 and D12 ≥ 0) are called the nonnegativity constraints (or nonnegativity conditions). 5. The other constraints are referred to as functional constraints (or structural constraints). 6. The parameters of the model are the constants in the algebraic model (the numbers in the data cells). 7. Any choice of values for the decision variables (regardless of how desirable or undesirable the choice) is called a solution for the model. 8. A feasible solution is one that satisfies all the constraints, whereas an infeasible ­solution violates at least one constraint. 9. The best feasible solution, the one that maximizes P (or G12), is called the optimal ­solution. (It is possible to have a tie for the best feasible solution, in which case all the tied solutions are called optimal solutions.) Comparisons So what are the relative advantages of algebraic models and spreadsheet models? An algebraic model provides a very concise and explicit statement of the problem. Sophisticated software packages that can solve huge problems generally are based on algebraic models because of both their compactness and their ease of use in rescaling the size of a problem. Management 2.4 The Graphical Method for Solving Two-Variable Problems 35 Management scientists often use algebraic models, but managers generally prefer spreadsheet models. science practitioners with an extensive mathematical background find algebraic models very useful. For others, however, spreadsheet models are far more intuitive. Both managers and business students training to be managers generally live with spreadsheets, not algebraic models. Therefore, the emphasis throughout this book is on spreadsheet models. Review Questions 1. When formulating a linear programming model, what are the initial steps that are the same with either a spreadsheet formulation or an algebraic formulation? 2. When formulating a linear programming model algebraically, algebraic symbols need to be introduced to represent which kinds of quantities in the model? 3. What are decision variables for a linear programming model? The objective function? Nonnegativity constraints? Functional constraints? 4. What is meant by a feasible solution for the model? An optimal solution? 2.4 THE GRAPHICAL METHOD FOR SOLVING TWO-VARIABLE PROBLEMS graphical method The graphical method provides helpful intuition about linear programming. Linear programming problems having only two decision variables, like the Wyndor problem, can be solved by a graphical method. Although this method cannot be used to solve problems with more than two decision variables (and most linear programming problems have far more than two), it still is well worth learning. The procedure provides geometric intuition about what linear programming is and what it is trying to achieve. This intuition is very helpful in analyzing linear programming models, including larger problems that cannot be solved directly by the graphical method. For example, it enables visualizing what is happening when performing what-if analysis for linear programming (the subject of Chapter 5). It is more convenient to apply the graphical method to the algebraic version of the linear programming model rather than the spreadsheet version. We shall briefly illustrate the method by using the algebraic model obtained for the Wyndor problem in the preceding section. (A far more detailed description of the graphical method, including its application to the Wyndor problem, is provided in the supplement to this chapter that is available at www.mhhe.com/Hillier6e.) For this purpose, keep in mind that D = Production rate for the special new doors (the number in changing cell C12 of the spreadsheet) W = Production rate for the special new windows (the number in changing cell D12 of the spreadsheet) The key to the graphical method is the fact that possible solutions can be displayed as points on a two-dimensional graph that has a horizontal axis giving the value of D and a vertical axis giving the value of W. Figure 2.5 shows some sample points. Notation: Either (D, W) = (2, 3) or just (2, 3) refers to the solution where D = 2 and W = 3, as well as to the corresponding point in the graph. Similarly, (D, W) = (4, 6) means D = 4 and W = 6, whereas the origin (0, 0) means D = 0 and W = 0. feasible region The points in the feasible region are those that satisfy every constraint. To find the optimal solution (the best feasible solution), we first need to display graphically where the feasible solutions are. To do this, we must consider each constraint, identify the solutions graphically that are permitted by that constraint, and then combine this information to identify the solutions permitted by all the constraints. The solutions permitted by all the constraints are the feasible solutions and the portion of the two-dimensional graph where the feasible solutions lie is referred to as the feasible region. The shaded region in Figure 2.6 shows the feasible region for the Wyndor problem. We now will outline how this feasible region was identified by considering the five constraints one at a time. To begin, the constraint D ≥ 0 implies that consideration must be limited to points that lie on or to the right of the W axis. Similarly, the constraint W ≥ 0 restricts consideration to the points on or above the D axis. 36 Chapter Two Linear Programming: Basic Concepts FIGURE 2.5 W Graph showing the points (D, W) = (2, 3) and (D, W) = (4, 6) for the Wyndor Glass Co. product-mix problem. Production Rate (units per week) for Windows 8 –1 (4, 6) 6 5 4 A product mix of D = 2 and W = 3 (2, 3) 3 2 1 –2 A product mix of D = 4 and W = 6 7 Origin 0 1 –1 2 3 4 5 6 7 Production Rate (units per week) for Doors 8 D –2 Next, consider the first functional constraint, D ≤ 4, which limits the usage of Plant 1 for producing the special new doors to a maximum of four hours per week. The solutions permitted by this constraint are those that lie on, or to the left of, the vertical line that intercepts the D axis at D = 4, as indicated by the arrows pointing to the left from this line in Figure 2.6. FIGURE 2.6 W 10 Production Rate for Windows Graph showing how the feasible region is formed by the constraint boundary lines, where the arrows indicate which side of each line is permitted by the corresponding constraint. 8 3D + 2W = 18 D=4 2W = 12 6 4 Feasible region 2 0 2 4 6 Production Rate for Doors 8 D 2.4 The Graphical Method for Solving Two-Variable Problems 37 For any constraint with an inequality sign, its constraint boundary equation is obtained by replacing the inequality sign by an equality sign. The second functional constraint, 2W ≤ 12, has a similar effect, except now the boundary of its permissible region is given by a horizontal line with the equation, 2W = 12 (or W = 6), as indicated by the arrows pointing downward from this line in Figure 2.6. The line forming the boundary of what is permitted by a constraint is sometimes referred to as a constraint boundary line, and its equation may be called a constraint boundary equation. Frequently, a constraint boundary line is identified by its equation. For each of the first two functional constraints, D ≤ 4 and 2W ≤ 12, note that the equation for the constraint boundary line (D = 4 and 2W = 12, respectively) is obtained by replacing the inequality sign with an equality sign. For any constraint with an inequality sign (whether a functional constraint or a nonnegativity constraint), the general rule for obtaining its constraint boundary equation is to substitute an equality sign for the inequality sign. We now need to consider one more functional constraint, 3D + 2W ≤ 18. Its constraint boundary equation 3D + 2W = 18 includes both variables, so the boundary line it represents is neither a vertical line nor a horizontal line. Therefore, the boundary line must intercept (cross through) both axes somewhere. But where? When a constraint boundary line is neither a vertical line nor a horizontal line, the line intercepts the D axis at the point on the line where W = 0. Similarly, the line intercepts the W axis at the point on the line where D = 0. Hence, the constraint boundary line 3D + 2W = 18 intercepts the D axis at the point where W = 0. When W = 0, 3D + 2W = 18 becomes 3D = 18 so the intercept with the D axis is at D= 6 Similarly, the line intercepts the W axis where D = 0. When D = 0, 3D + 2W = 18 becomes 2W = 18 so the intercept with the D axis is at W= 9 The location of a slanting constraint boundary line is found by identifying where it intercepts each of the two axes. Checking whether (0, 0) satisfies a constraint indicates which side of the constraint boundary line satisfies the constraint. Consequently, the constraint boundary line is the line that passes through these two intercept points, as shown in Figure 2.6. As indicated by the arrows emanating from this line in Figure 2.6, the solutions permitted by the constraint 3D + 2W ≤ 18 are those that lie on the origin side of the constraint boundary line 3D + 2W = 18. The easiest way to verify this is to check whether the origin itself, (D, W) = (0, 0), satisfies the constraint.3 If it does, then the permissible region lies on the side of the constraint boundary line where the origin is. Otherwise, it lies on the other side. In this case, 3(0) + 2(0) = 0 so (D, W) = (0, 0) satisfies 3D + 2W ≤ 18 (In fact, the origin satisfies any constraint with a ≤ sign and a positive right-hand side.) A feasible solution for a linear programming problem must satisfy all the constraints simultaneously. The arrows in Figure 2.6 indicate that the nonnegative solutions permitted by each of these constraints lie on the side of the constraint boundary line where the origin is (or actually on the line itself). Therefore, the feasible solutions are the nonnegative solutions that lie nearer to the origin than all three constraint boundary lines (or actually on the line nearest the origin at that point). Having identified the feasible region, the final step is to find which of these feasible solutions is the best one—the optimal solution. For the Wyndor problem, the objective happens to 3 The one case where using the origin to help determine the permissible region does not work is if the constraint boundary line passes through the origin. In this case, any other point not lying on this line can be used just like the origin. 38 Chapter Two Linear Programming: Basic Concepts be to maximize the total profit per week from the two products (denoted by P). Therefore, we want to find the feasible solution (D, W) that makes the value of the objective function P = 300D + 500W as large as possible. To accomplish this, we need to be able to locate all the points (D, W) on the graph that give a specified value of the objective function. For example, consider a value of P = 1,500 for the objective function. Which points (D, W) give 300D + 500W = 1,500? This equation is the equation of a line. Just as when plotting constraint boundary lines, the location of this line is found by identifying its intercepts with the two axes. When W = 0, this equation yields D = 5, and similarly, W = 3 when D = 0, so these are the two intercepts, as shown by the bottom slanting line passing through the feasible region in Figure 2.7. P = 1,500 is just one sample value of the objective function. For any other specified value of P, the points (D, W) that give this value of P also lie on a line called an objective function line. An objective function line is a line whose points all have the same value of the objective function. For the bottom objective function line in Figure 2.7, the points on this line that lie in the feasible region provide alternate ways of achieving an objective function value of P = 1,500. Can we do better? Let us try doubling the value of P to P = 3,000. The corresponding objective function line 300D + 500W = 3,000 is shown as the middle line in Figure 2.7. (Ignore the top line for the moment.) Once again, this line includes points in the feasible region, so P = 3,000 is achievable. Let us pause to note two interesting features of these objective function lines for P = 1,500 and P = 3,000. First, these lines are parallel. Second, doubling the value of P from 1,500 to 3,000 also doubles the value of W at which the line intercepts the W axis from W = 3 to W = 6. These features are no coincidence, as indicated by the following properties. Key Properties of Objective Function Lines: All objective function lines for the same problem are parallel. Furthermore, the value of W at which an objective function line intercepts the W axis is proportional to the value of P. FIGURE 2.7 Graph showing three objective function lines for the Wyndor Glass Co. product-mix problem, where the top one passes through the optimal solution. Production Rate W for Windows 8 P = 3,600 = 300D + 500W Optimal solution P = 3,000 = 300D + 500W 6 4 P = 1,500 = 300D + 500W (2, 6) Feasible region 2 0 2 4 6 Production Rate for Doors 8 10 D 2.5 Using Excel’s Solver to Solve Linear Programming Problems 39 These key properties of objective function lines suggest the strategy to follow to find the optimal solution. We already have tried P = 1,500 and P = 3,000 in Figure 2.7 and found that their objective function lines include points in the feasible region. Increasing P again will generate another parallel objective function line farther from the origin. The objective function line of special interest is the one farthest from the origin that still includes a point in the feasible region. This is the third objective function line in Figure 2.7. The point on this line that is in the feasible region, (D, W) = (2, 6), is the optimal solution since no other feasible solution has a larger value of P. Optimal Solution D = 2 (Produce 2 special new doors per week) W = 6 (Produce 6 special new windows per week) These values of D and W can be substituted into the objective function to find the value of P. P = 300D + 500W = 300(2) + 500(6) = 3,600 This has been a fairly quick description of the graphical method. You can go to the supplement to this chapter that is available at www.mhhe.com/Hillier6e if you would like to see a fuller description of how to use the graphical method to solve the Wyndor problem. The end of Section 2.7 will provide another example of the graphical method, where the objective in that case is to minimize instead of maximize the objective function. An additional example also is included in the first of this chapter’s solved problems (Problem 2.S-1) presented at the end of the chapter. Summary of the Graphical Method The graphical method can be used to solve any linear programming problem having only two decision variables. The method uses the following steps: 1. Draw the constraint boundary line for each functional constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint. 2. Find the feasible region by determining where all constraints are satisfied simultaneously. 3. Determine the slope of one objective function line. All other objective function lines will have the same slope. 4. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line. 5. A feasible point on the optimal objective function line is an optimal solution. Review Questions 1. The graphical method can be used to solve linear programming problems with how many decision variables? 2. What do the axes represent when applying the graphical method to the Wyndor problem? 3. What is a constraint boundary line? A constraint boundary equation? 4. What is the easiest way of determining which side of a constraint boundary line is permitted by the constraint? 2.5 USING EXCEL’S SOLVER TO SOLVE LINEAR PROGRAMMING PROBLEMS The graphical method is very useful for gaining geometric intuition about linear programming, but its practical use is severely limited by only being able to solve tiny problems with two decision variables. Another procedure that will solve linear programming problems of any reasonable size is needed. Fortunately, Excel includes a tool called Solver that will do this once the spreadsheet model has been formulated as described in Section 2.2. (Section 2.6 will show how the Analytic Solver software package, which includes a more advanced version of Solver, can be used to solve this same problem.) To access Solver the first time, you need to 40 Chapter Two Linear Programming: Basic Concepts Excel Tip: If you select cells by clicking on them, they will first appear in the dialog box with their cell addresses and with dollar signs (e.g., $C$9:$D$9). You can ignore the dollar signs. Solver eventually will replace both the cell addresses and the dollar signs with the corresponding range name (if a range name has been defined for the given cell addresses), but only after either adding a constraint or closing and reopening the Solver dialog box. Solver Tip: To select changing cells, click and drag across the range of cells. If the changing cells are not contiguous, you can type a comma and then select another range of cells. Up to 200 changing cells can be selected with the basic version of Solver that comes with Excel. FIGURE 2.8 The incomplete Solver dialog box after specifying the first components of the model for the ­Wyndor problem. TotalProfit (G12) is being maximized by changing UnitsProduced (C12:D12). Figure 2.9 will demonstrate the addition of constraints and then Figure 2.10 will demonstrate the changes needed to specify that the problem being considered is a linear programming problem. (The default Solving Method shown here—GRG Nonlinear— is not applicable to linear programming problems.) install it. In Windows versions of Excel, choose Excel Options from the File menu, then click on Add-Ins on the left side of the window, select Manage Excel Add-Ins at the bottom of the window, and then press the Go button. Make sure Solver is selected in the Add-Ins dialog box, and then it should appear on the Data tab. For Mac versions of Excel, select Add-Ins from the Tools menu, make sure Solver is selected, then click OK; Solver should now appear on the Data tab. Figure 2.3 in Section 2.2 shows the spreadsheet model for the Wyndor problem. The values of the decision variables (the production rates for the two products) are in the changing cells, UnitsProduced (C12:D12), and the value of the objective function (the total profit per week from the two products) is in the objective cell TotalProfit (G12). To get started, an arbitrary trial solution has been entered by placing zeroes in the changing cells. Solver will then change these to the optimal values after solving the problem. This procedure is started by choosing Solver on the Data tab. Figure 2.8 shows the Solver dialog box that is used to tell Solver where each component of the model is located on the spreadsheet. You have the choice of typing the range names, typing the cell addresses, or probably the easiest method, clicking on the cells in the spreadsheet. Figure 2.8 shows the results of the last method—clicking on the cell for TotalProfit (G12) for the objective cell. Since the goal is to maximize the objective cell, Max also has been selected. The next entry in the Solver dialog box identifies the changing cells, specified by selecting the cells for UnitsProduced (C12:D12). Next, the cells containing the functional constraints need to be specified. This is done by clicking on the Add button on the Solver dialog box. This brings up the Add Constraint dialog box shown in Figure 2.9. The ≤ signs in cells F7, F8, and F9 of Figure 2.3 are a reminder that the cells in HoursUsed (E7:E9) all need to be less than or equal to the corresponding cells in HoursAvailable (G7:G9). These constraints are specified for Solver by 2.5 Using Excel’s Solver to Solve Linear Programming Problems 41 The Add Constraint dialog box is used to specify all the functional constraints. When solving a linear programming problem, be sure to specify that nonnegativity constraints are needed and that the model is linear by choosing Simplex LP. FIGURE 2.9 The Add Constraint dialog box after specifying that cells E7, E8, and E9 in Figure 2.3 are required to be less than or equal to cells G7, G8, and G9, respectively. FIGURE 2.10 The completed Solver dialog box after specifying the entire model in terms of the spreadsheet. entering HoursUsed (or E7:E9) on the left-hand side of the Add Constraint dialog box and then entering HoursAvailable (or G7:G9) on the right-hand side. For the sign between these two sides, there is a menu to choose between < =, =, or > =, so < = has been chosen. This choice is needed even though ≤ signs were previously entered in column F of the spreadsheet because the Solver only uses the constraints that are specified with the Add Constraint dialog box. If there were more functional constraints to add, you would click on Add to bring up a new Add Constraint dialog box. However, since there are no more in this example, the next step is to click on OK to go back to the Solver dialog box. Before asking Solver to solve the model, two more steps need to be taken. We need to tell Solver that nonnegativity constraints are needed for the changing cells to reject negative production rates. We also need to specify that this is a linear programming problem so the simplex method (the standard method used by Solver to solve linear programming problems) can be used. This is demonstrated in Figure 2.10, where the Make Unconstrained 42 Chapter Two Linear Programming: Basic Concepts When you define range names for cells in your spreadsheet, these range names are displayed by Solver. This makes the model in Solver much easier to interpret. Solver Tip: The message “Solver could not find a feasible solution” means that there are no solutions that satisfy all the constraints. The message “The Objective Cell values do not converge” means that Solver could not find a best solution, because better solutions always are available (e.g., if the constraints do not prevent infinite profit). The message “The linearity conditions required by this LP Solver are not satisfied” means Simplex LP was chosen as the Solving Method, but the model is not linear. FIGURE 2.11 The Solver Results dialog box that indicates that an optimal solution has been found. Variables Non-Negative option has been checked and the Solving Method chosen is Simplex LP (rather than GRG Nonlinear or Evolutionary, which are used for solving nonlinear problems). The Solver dialog box shown in this figure now summarizes the complete model. Note that after returning from the Add Constraint dialog box, Solver has now converted all of the cell addresses to the corresponding range names that were defined for these cells (TotalProfit for G12, UnitsProduced for C12:D12, HoursUsed for E7:E9, and HoursAvailable for G7:G9). Defining range names makes the model in Solver much easier to interpret. Now you are ready to click on Solve in the Solver dialog box, which will start the solving of the problem in the background. After a few seconds (for a small problem), Solver will then indicate the results. Typically, it will indicate that it has found an optimal solution, as specified in the Solver Results dialog box shown in Figure 2.11. If the model has no feasible solutions or no optimal solution, the dialog box will indicate that instead by stating that “Solver could not find a feasible solution” or that “The Objective Cell values do not converge.” (Section 14.1 will describe how these possibilities can occur.) The dialog box also presents the option of generating various reports. One of these (the Sensitivity Report) will be discussed in detail in Chapter 5. After solving the model and clicking OK in the Solver Results dialog box, Solver replaces the original numbers in the changing cells with the optimal numbers, as shown in Figure 2.12. Thus, the optimal solution is to produce two doors per week and six windows per week, just as was found by the graphical method in the preceding section. The spreadsheet also indicates the corresponding number in the objective cell (a total profit of $3,600 per week), as well as the numbers in the output cells HoursUsed (E7:E9). The entries needed for the Solver dialog box are summarized in the Solver Parameters box shown on the bottom left of Figure 2.12. This more compact summary of the Solver Parameters will be shown for all of the many models that involve the Solver throughout the book. At this point, you might want to check what would happen to the optimal solution if any of the numbers in the data cells were to be changed to other possible values. This is easy to do because Solver saves all the addresses for the objective cell, changing cells, constraints, and so on when you save the file. All you need to do is make the changes you want in the data cells and then click on Solve in the Solver dialog box again. (Chapter 5 will focus on this kind of what-if analysis, including how to use the Solver’s Sensitivity Report to expedite the analysis.) To assist you with experimenting with these kinds of changes, www.mhhe.com/Hillier6e includes Excel files for this chapter (as for others) that provide a complete formulation and 2.6 Analytic Solver 43 FIGURE 2.12 The spreadsheet obtained after solving the Wyndor problem. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Doors Windows $300 $500 Unit Profit 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 2 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 2 6 $3,600 10 11 12 Units Produced Solver Parameters Set Objective Cell: Total Profit To: Max By Changing Variable Cells: UnitsProduced Subject to the Constraints: HoursUsed <= HoursAvailable Solver Options: Make Variables Nonnegative Solving Method: Simplex LP E 5 Hours 6 7 8 9 Used =SUMPRODUCT(C7:D7, UnitsProduced) =SUMPRODUCT(C8:D8, UnitsProduced) =SUMPRODUCT(C9:D9, UnitsProduced) G 11 12 Total Profit =SUMPRODUCT(UnitProfit, UnitsProduced) Range Name HoursAvailable HoursUsed HoursUsedPerUnitProduced TotalProfit UnitProfit UnitsProduced Cell G7:G9 E7:E9 C7:D9 G12 C4:D4 C12:D12 solution of the examples here (the Wyndor problem and the one in Section 2.7) in a spreadsheet format. We encourage you to “play” with these examples to see what happens with different data, different solutions, and so forth. You might also find these spreadsheets useful as templates for homework problems. Review Questions 2.6 1. Which dialog box is used to enter the addresses for the objective cell and the changing cells? 2. Which dialog box is used to specify the functional constraints for the model? 3. Which options normally need to be chosen to solve a linear programming model? ANALYTIC SOLVER Frontline Systems, the original developer of the standard Solver included with Excel (hereafter referred to as Excel’s Solver), also has developed Premium versions of Solver that provide greatly enhanced functionality. The company now features a particularly powerful Premium Solver called Analytic Solver. We are excited to provide access to the Excel add-in, Analytic Solver for Education, which includes all the features of the full-fledged Analytic Solver (also called Analytic Solver Comprehensive) except that it is limited to considerably smaller problems than the huge problems that the full-fledged version can solve. However, since Analytic Solver for Education can handle much larger problems than any problems you will encounter in this book, we will simply refer to it hereafter as Analytic Solver. 44 Chapter Two Linear Programming: Basic Concepts Instructions for obtaining a low-priced student license to download this software are provided in both the preface and www.mhhe.com/Hillier6e. A cloud-based version of this software is also available at AnalyticSolver.com. The cloud-based version works using a browser, but is designed to look and feel as much as possible like the downloaded Analytic Solver add-in for Excel. While Excel’s Solver is sufficient for most of the problems considered in this book, Analytic Solver includes a number of important features not available with Excel’s Solver. Where either Excel’s Solver or Analytic Solver can be used, the book will often use the term Solver generically to mean either Excel’s Solver or Analytic Solver. Where there are differences, the book will include instructions for both Excel’s Solver and Analytic Solver. The enhanced features of Analytic Solver will be highlighted as they come up throughout the book. However, if you and your instructor prefer to focus on only using Excel’s Solver, you will find that there is plenty of material to cover in the book that does not require the use of Analytic Solver. In this case, you can skip this section and continue on without loss of continuity. When Analytic Solver is installed, two new tabs are available on the Excel ribbon named Analytic Solver and Data Mining (along with an optional third tab named Solver Home). Choosing the Analytic Solver tab will reveal the ribbon shown in Figure 2.13. (The Data ­Mining tab will not be used in this chapter, but will be introduced in Chapter 10.) The buttons on this ribbon will be used to interact with Analytic Solver. This same figure also reveals a nice feature of Analytic Solver—the Solver Options and Model Specifications pane (showing the objective cell, changing cells, constraints, etc.)—that can be seen alongside your main spreadsheet, with both visible simultaneously. This pane can be toggled on (to see the model) or off (to hide the model and leave more room for the spreadsheet) by clicking on the Model button on the far left of the Analytic Solver ribbon. Also, since the model was already set up with Excel’s Solver in Section 2.5, it is already set up in the Model pane, with the objective specified as TotalProfit (G12) with changing cells UnitsProduced (C12:D12) and the constraints HoursUsed (E7:E9) <= HoursAvailable (G7:G9). The data for Excel’s Solver and Analytic Solver are compatible with each other. Making a change with one makes the same change in the other. Thus, you can work with either Excel’s Solver or Analytic Solver, and then go back and forth, without losing any Solver data. If the model had not been previously set up with Excel’s Solver, the steps for doing so with Analytic Solver are analogous to the steps used with Excel’s Solver as covered in Section 2.5. In both cases, we need to specify the location of the objective cell, the changing cells, and the functional constraints, and then click to solve the model. However, the user interface is somewhat different. Analytic Solver uses the buttons on the Analytic Solver ribbon instead of the Solver dialog box. We will now walk you through the steps to set up the Wyndor problem in Analytic Solver. To specify TotalProfit (G12) as the objective cell, select the cell in the spreadsheet and then click on the Objective button on the Analytic Solver ribbon. As shown in Figure 2.14, FIGURE 2.13 The screenshot for the Wyndor problem that shows both the ribbon and the Solver Options and Model Specifications pane that are revealed after choosing the tab called Analytic Solver on the Excel ribbon. 2.6 Analytic Solver 45 FIGURE 2.14 The screenshot for the Wyndor problem that shows the drop-down menu generated by clicking on the Objective button on the Analytic Solver ribbon after choosing TotalProfit (G12) as the objective cell. Analytic Solver Tip: Another way to add an objective, changing cells, or constraints using Analytic Solver is to click on the big green plus (+) on the Mode pane and choose Add Objective, Add Variable, or Add Constraint, respectively. FIGURE 2.15 The screenshot for the Wyndor problem that shows the drop-down menu generated by clicking on the Decision button on the Analytic Solver ribbon after choosing UnitsProduced (C12:D12) as the changing variable cells. FIGURE 2.16 The screenshot of the Wyndor problem that shows the drop-down menu generated by clicking on the Constraints button on the Analytic Solver ribbon after choosing HoursUsed (E7:E9) as the left-hand side of the functional constraints. this will drop down a menu where you can choose to minimize (Min) or maximize (Max) the objective cell. Within the options of Min or Max are further options (Normal, Expected, Chance). For now, we will always choose the Normal option. To specify UnitsProduced (C12:D12) as the changing cells, select these cells in the spreadsheet and then click on the Decisions button on the Analytic Solver ribbon. As shown in Figure 2.15, this will drop down a menu where you can choose various options (Plot, Normal, Recourse). For linear programming, we will always choose the Normal option. Next the functional constraints need to be specified. For the Wyndor problem, the functional constraints are HoursUsed (E7:E9) <= HoursAvailable (G7:G9). To enter these constraints in Analytic Solver, select the cells representing the left-hand side of these constraints (HoursUsed, or E7:E9) and click the Constraints button on the Analytic Solver ribbon. As shown in Figure 2.16, this drops down a menu for various kinds of constraints. For linear programming functional 46 Chapter Two Linear Programming: Basic Concepts FIGURE 2.17 The Add Constraint dialog box that is brought up after choosing <= in Figure 2.16. This dialog box also shows HoursAvailable (G7:G9) as the righthand side of the functional constraints after clicking in the Constraint box and choosing these cells on the spreadsheet. Analytic Solver Tip: Double-clicking on any element of the model (e.g., the objective, any of the changing cells, or any of the constraints) will bring up a dialog box allowing you to make changes to that part of the model. FIGURE 2.18 This figure shows the result of clicking on the HoursUsed <= Hours Available constraint in the Model pane prior to considering possible changes in the constraint. constraints, choose Normal Constraint and then the type of constraint desired (either <=, =, or >=). For the Wyndor problem, choosing <= would then bring up the Add Constraint dialog box shown in Figure 2.17. This is much like the Add Constraint dialog box for Excel’s Solver (see ­Figure 2.9). HoursUsed and <= already are filled in when the Add Constraint dialog box is brought up (because the HoursUsed cells were selected and <= was chosen under the Constraints button menu). The Add Constraint dialog box then can be used to fill in the remaining right-hand side of the constraint—HoursAvailable (G7:G9)—by clicking in the box labeled Constraint and choosing these cells on the spreadsheet. Figure 2.17 shows the dialog box after this has been done. Changes to the model can easily be made within the Model pane shown in Figure 2.13. For example, to delete an element of the model (e.g., the objective, changing cells, or constraints), select that part of the model and then click on the red X near the top of the Model pane. To change an element of the model, click on that element in the Model pane. The bottom of the Model pane will then show information about that element. For example, clicking on the Hours Used <= HoursAvailable constraint in the Model pane will then show the information seen in Figure 2.18. Clicking on any piece of the information will allow you to change it (e.g., you can change <= to >=, or you can change the cell references for either side of the constraint). Selecting the Engine tab at the top of the Model pane will show information about the algorithm that will be used to solve the problem as well as a variety of options for that algorithm. The drop-down menu at the top will allow you to choose the algorithm. For a linear programming model (such as the Wyndor problem), you will want to choose the Standard LP/Quadratic Engine. This is equivalent to the Simplex LP option in Excel’s Solver. To make unconstrained variables nonnegative (as we did in Figure 2.10 with Excel’s Solver), be sure that the Assume Nonnegative option is set to true. Figure 2.19 shows the model pane after making these selections. 2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 47 FIGURE 2.19 The Engine tab of the Model pane after selecting the Standard LP/­Quadratic Engine and setting the Assume Nonnegative option to True. FIGURE 2.20 A screenshot showing the final solution of the Wyndor problem and the Output tab of the model pane showing a summary of the solution process. Once the model is all set up in Analytic Solver, the model would be solved by clicking on the Optimize button on the Analytic Solver ribbon or the green arrow button on the Model pane. Just like Excel’s Solver, this will then display the results of solving the model on the spreadsheet, as shown in Figure 2.20. As seen in this figure, the Output tab of the Model pane also will show a summary of the solution process, including the message (similar to Figure 2.11) that “Solver found a solution. All constraints and optimality conditions are satisfied.” Review Questions 1. 2. 3. 4. Which button on the Analytic Solver ribbon should be pressed to specify the objective cell? Which button on the Analytic Solver ribbon should be pressed to specify the changing cells? Which button on the Analytic Solver ribbon should be pressed to enter the constraints? Which button on the Analytic Solver ribbon should be pressed to solve the model? 2.7 A MINIMIZATION EXAMPLE—THE PROFIT & GAMBIT CO. ADVERTISING-MIX PROBLEM The analysis of the Wyndor Glass Co. case study in Sections 2.2, 2.5, and 2.6 illustrated how to formulate and solve one type of linear programming model on a spreadsheet. The same general approach can be applied to many other problems as well. The great flexibility of linear programming and spreadsheets provides a variety of options for how to adapt the 48 Chapter Two Linear Programming: Basic Concepts formulation of the spreadsheet model to fit each new problem. Our next example illustrates some options not used for the Wyndor problem. Planning an Advertising Campaign The Profit & Gambit Co. produces cleaning products for home use. This is a highly competitive market, and the company continually struggles to increase its small market share. Management has decided to undertake a major new advertising campaign that will focus on the following three key products: • A spray prewash stain remover. • A liquid laundry detergent. • A powder laundry detergent. This campaign will use both television and the print media. A commercial has been developed to run on national television that will feature the liquid detergent. The advertisement for the print media will promote all three products and will include cents-off coupons that consumers can use to purchase the products at reduced prices. The general goal is to increase the sales of each of these products (but especially the liquid detergent) over the next year by a significant percentage over the past year. Specifically, management has set the following goals for the campaign: • Sales of the stain remover should increase by at least 3 percent. • Sales of the liquid detergent should increase by at least 18 percent. • Sales of the powder detergent should increase by at least 4 percent. Table 2.2 shows the estimated increase in sales for each unit of advertising in the respective outlets.4 (A unit is a standard block of advertising that Profit & Gambit commonly purchases, but other amounts also are allowed.) The reason for −1 percent for the powder detergent in the Television column is that the TV commercial featuring the new liquid detergent will take away some sales from the powder detergent. The bottom row of the table shows the cost per unit of advertising for each of the two outlets. Management’s objective is to determine how much to advertise in each medium to meet the sales goals at a minimum total cost. Formulating a Spreadsheet Model for This Problem The procedure summarized at the end of Section 2.2 can be used to formulate the spreadsheet model for this problem. Each step of the procedure is repeated below, followed by a description of how it is performed here. 1. Gather the data for the problem. This has been done as presented in Table 2.2. 2. Enter the data into data cells on a spreadsheet. The top half of Figure 2.21 shows this spreadsheet. The data cells are in columns C and D (rows 4 and 8 to 10), as well as in cells G8:G10. Note how this particular formatting of the spreadsheet has facilitated a direct transfer of the data from Table 2.2. TABLE 2.2 Increase in Sales per Unit of Advertising Data for the Profit & Gambit Co. AdvertisingMix Problem Product Stain remover Liquid detergent Powder detergent Unit cost Television 0% 3 –1 $1 million Print Media 1% 2 4 Minimum Required Increase 3% 18 4 $2 million 4 A simplifying assumption is being made that each additional unit of advertising in a particular outlet will yield the same increase in sales regardless of how much advertising already is being done. This becomes a poor assumption when the levels of advertising under consideration can reach a saturation level (as in Case 8.1), but is a reasonable approximation for the small levels of advertising being considered in this problem. 2.7 A Minimization Example—The Profit & Gambit Co. Advertising-Mix Problem 49 FIGURE 2.21 The spreadsheet model for the Profit & Gambit problem, including the formulas for the objective cell TotalCost (G14) and the other output cells in column E, as well as the specifications needed to set up Solver. The changing cells, AdvertisingUnits (C14:D14), show the optimal solution obtained by Solver. A 1 B C D E F G Profit & Gambit Co. Advertising-Mix Problem 2 3 4 Unit Cost ($millions) Television Print Media 1 2 5 6 Increased Minimum Sales Increase Increase in Sales per Unit of Advertising 7 8 Stain Remover 0% 1% 3% ≥ 3% 9 Liquid Detergent 3% 2% 18% ≥ 18% 10 Powder Detergent –1% 4% 8% ≥ 4% 11 12 Total Cost 13 14 Advertising Units Solver Parameters Set Objective Cell: TotalCost To: Min By Changing (Variable) Cells: AdvertisingUnits Subject to the Constraints: IncreasedSales >= MinimumIncrease Solver Options: Make Variables Nonnegative Solving Method: Simplex LP Television Print Media ($millions) 4 3 10 E 6 Increased 7 8 9 10 Sales =SUMPRODUCT(C8:D8, AdvertisingUnits) =SUMPRODUCT(C9:D9, AdvertisingUnits) =SUMPRODUCT(C10:D10, AdvertisingUnits) G 12 Total Cost 13 ($millions) 14 =SUMPRODUCT(UnitCost, AdvertisingUnits) Range Name AdvertisingUnits IncreasedSales IncreasedSalesPerUnitAdvertising MinimumIncrease TotalCost UnitCost Cells C14: D14 E8: E10 C8: D10 G8: G10 G14 C4: D4 3. Identify the decisions to be made on the levels of activities and designate changing cells for making these decisions. In this case, the activities of concern are advertising on television and advertising in the print media, so the levels of these activities refer to the amount of advertising in these media. Therefore, the decisions to be made are Decision 1: TV = Number of units of advertising on television Decision 2: PM = Number of units of advertising in the print media The two gray cells with light borders in Figure 2.21—C14 and D14—have been designated as the changing cells to hold these numbers: TV → cell C14 PM → cell D14 with AdvertisingUnits as the range name for these cells. (See the bottom of Figure 2.21 for a list of all the range names.) These are natural locations for the changing cells, since each one is in the column for the corresponding advertising medium. To get started, an arbitrary trial solution (such as all zeroes) is entered into these cells. (Figure 2.21 shows the optimal solution after having already applied Solver.) 50 Chapter Two Linear Programming: Basic Concepts 4. Identify the constraints on these decisions and introduce output cells as needed to specify these constraints. The three constraints imposed by management are the goals for the increased sales for the respective products, as shown in the rightmost column of Table 2.2. These constraints are Unlike the Wyndor problem, we need to use ≥ signs for these constraints. Stain remover: Total increase in sales ≥ 3% detergent: Liquid Total increase in sales ≥ 18% Powder detergent: Total increase in sales ≥ 4% The second and third columns of Table 2.2 indicate that the total increases in sales from both forms of advertising are Total for stain remover = 1% of PM Total for liquid detergent = 3% of TV + 2% of PM Total for powder detergent = − 1% of TV + 4% of PM Unlike the Wyndor problem, the objective now is to minimize the objective cell. Consequently, since rows 8, 9, and 10 in the spreadsheet are being used to provide information about the three products, cells E8, E9, and E10 are introduced as output cells to show the total increase in sales for the respective products. In addition, ≥ signs have been entered in column F to remind us that the increased sales need to be at least as large as the numbers in column G. (The use of ≥ signs here rather than ≤ signs is one key difference from the spreadsheet model for the Wyndor problem in Figure 2.3.) 5. Choose the overall measure of performance to be entered into the objective cell. Management’s stated objective is to determine how much to advertise in each medium to meet the sales goals at a minimum total cost. Therefore, the total cost of the advertising is entered in the objective cell TotalCost (G14). G14 is a natural location for this cell since it is in the same row as the changing cells. The bottom row of Table 2.2 indicates that the number going into this cell is Cost = ( $1 million) TV + ( $2 million) PM → cell G14 6. Use a SUMPRODUCT function to enter the appropriate value into each output cell (including the objective cell). Based on the above expressions for cost and total increases in sales, the SUMPRODUCT functions needed here for the output cells are those shown under the right side of the spreadsheet in Figure 2.21. Note that each of these functions involves the relevant data cells and the changing cells, AdvertisingUnits (C14:D14). This spreadsheet model is a linear programming model, since it possesses all the characteristics of such models enumerated in Section 2.2. Applying Solver to This Model The procedure for using Solver to obtain an optimal solution for this model is basically the same as described in Section 2.5 (for Excel’s Solver) or Section 2.6 (for Analytic Solver). The Solver parameters are shown below the left-hand side of the spreadsheet in Figure 2.21. In addition to specifying the objective cell and changing cells, the constraints that IncreasedSales ≥ MinimumIncrease have been specified in this box by using the Add Constraint dialog box. Since the objective is to minimize total cost, Min also has been selected. (This is in contrast to the choice of Max for the Wyndor problem.) Two options are also specified at the bottom of the Solver Parameters box on the lower left-hand side of Figure 2.21. The changing cells need nonnegativity constraints (specified in the main Solver dialog box in Excel’s Solver, or on the Engine tab of the Model pane in Analytic Solver) because negative values of advertising levels are not possible alternatives. Choosing the Simplex LP solving method in Excel’s Solver (or the Standard LP/Quadratic Engine in Analytic Solver) specifies that this is a linear programming model. After running Solver, the optimal solution shown in the changing cells of the spreadsheet in Figure 2.21 is obtained. An Application Vignette Samsung Electronics Corp., Ltd. (SEC), is a leading merchant of dynamic and static random access memory devices and other advanced digital integrated circuits. Its site at Kiheung, South Korea (probably the largest semiconductor fabrication site in the world), fabricates more than 300,000 silicon wafers per month and employs over 10,000 people. Cycle time is the industry’s term for the elapsed time from the release of a batch of blank silicon wafers into the fabrication process until completion of the devices that are fabricated on those wafers. Reducing cycle times is an ongoing goal since it both decreases costs and enables offering shorter lead times to potential customers, a real key to maintaining or increasing market share in a very competitive industry. Three factors present particularly major challenges when striving to reduce cycle times. One is that the product mix changes continually. Another is that the company often needs to make substantial changes in the fab-out schedule inside the target cycle time as it revises forecasts of customer demand. The third is that the machines of a general type are not homogeneous so only a small number of machines are qualified to perform each device step. A management science team developed a huge linear programming model with tens of thousands of decision variables and functional constraints to cope with these challenges. The objective function involved minimizing back-orders and finished-goods inventory. The ongoing implementation of this model enabled the company to reduce manufacturing cycle times to fabricate dynamic random access memory devices from more than 80 days to less than 30 days. This tremendous improvement and the resulting reduction in both manufacturing costs and sale prices enabled Samsung to capture an additional $200 million in annual sales revenue. Source: R. C. Leachman, J. Kang, and Y. Lin, “SLIM: Short Cycle Time and Low Inventory in Manufacturing at Samsung Electronics,” Interfaces 32, no. 1 (January–February 2002), pp. 61–77. (A link to this article is available at www.mhhe.com/Hillier6e.) Optimal Solution C14 = 4 (Undertake 4 units of advertising on television) C14 = 3 (Undertake 3 units of advertising in the print media) The objective cell indicates that the total cost of this advertising plan would be $10 million. The Mathematical Model in the Spreadsheet When performing step 5 of the procedure for formulating a spreadsheet model, the total cost of advertising was determined to be Cost = TV + 2 PM (in millions of dollars) where the goal is to choose the values of TV (number of units of advertising on television) and PM (number of units of advertising in the print media) so as to minimize this cost. Step 4 identified three functional constraints: Stain remover: 1% of PM ≥ 3% detergent: Liquid 3% of TV + 2% of PM ≥ 18% Powder detergent: −1% of TV + 4% of PM ≥ 4% Choosing the Make Variables Non-Negative option with Solver recognized that TV and PM cannot be negative. Therefore, after dropping the percentage signs from the functional constraints, the complete mathematical model in the spreadsheet can be stated in the following succinct form. Minimize Cost = TV + 2 PM (in millions of dollars) subject to Stain remover increased sales: PM ≥ 3 Liquid detergent increased sales: 3 TV + 2 PM ≥ 18 Powder detergent increased sales: − TV + 4 PM ≥ 4 and TV ≥ 0 PM ≥ 0 Implicit in this statement is “Choose the values of TV and PM so as to . . . .” The term “subject to” is shorthand for “Choose these values subject to the requirement that the values satisfy all the following constraints.” 51 52 Chapter Two Linear Programming: Basic Concepts PM FIGURE 2.22 Graph showing two objective function lines for the Profit & Gambit Co. advertising-mix problem, where the bottom one passes through the optimal solution. 10 Feasible region Cost = 15 = TV + 2 PM Cost = 10 = TV + 2 PM 4 (4,3) optimal solution 0 10 5 Amount of TV advertising 15 TV This model is the algebraic version of the linear programming model in the spreadsheet (Figure 2.21). Note how the parameters (constants) of this algebraic model come directly from the numbers in Table 2.2. In fact, the entire model could have been formulated directly from this table. The differences between this algebraic model and the one obtained for the Wyndor problem in Section 2.3 lead to some interesting changes in how the graphical method is applied to solve the model. To further expand your geometric intuition about linear programming, we briefly describe this application of the graphical method next. Since this linear programming model has only two decision variables, it can be solved by the graphical method described in Section 2.4. The method needs to be adapted in two ways to fit this particular problem. First, because all the functional constraints now have a ≥ sign with a positive right-hand side, after obtaining the constraint boundary lines in the usual way, the arrows indicating which side of each line satisfies that constraint now all point away from the origin. Second, the method is adapted to minimization by moving the objective function lines in the direction that reduces Cost and then stopping at the last instant that an objective function line still passes through a point in the feasible region, where such a point then is an optimal solution. The supplement to this chapter includes a description of how the graphical method is applied to the Profit & Gambit problem in this way. Figure 2.22 shows the resulting final graph that identifies the optimal solution as TV = 4 (use 4 units of TV advertising) PM = 3 (use 3 units of print media advertising) Review Questions 2.8 1. 2. 3. 4. What kind of product is produced by the Profit & Gambit Co.? Which advertising media are being considered for the three products under consideration? What is management’s objective for the problem being addressed? What was the rationale for the placement of the objective cell and the changing cells in the spreadsheet model? 5. The algebraic form of the linear programming model for this problem differs from that for the Wyndor Glass Co. problem in which two major ways? LINEAR PROGRAMMING FROM A BROADER PERSPECTIVE Linear programming is an invaluable aid to managerial decision making in all kinds of companies throughout the world. The emergence of powerful spreadsheet packages has helped to further spread the use of this technique. The ease of formulating and solving small linear programming models on a spreadsheet now enables some managers with a very modest background in management science to do this themselves on their own desktop. 2.8 Linear Programming from a Broader Perspective 53 Many linear programming studies are major projects involving decisions on the levels of many hundreds or thousands of activities. For such studies, sophisticated software packages that go beyond spreadsheets generally are used for both the formulation and solution processes. These studies normally are conducted by technically trained teams of management scientists, sometimes called operations research analysts. Management instigates the study and then clarifies its objectives and needs, but the team does the rest. Consequently, there is little reason for a manager to know the details of how linear programming models are solved beyond the rudiments of using Solver. (Even most management science teams will use commercial software packages for solving their models on a computer rather than developing their own software.) Similarly, a manager does not need to know the technical details of how to formulate complex models, how to validate such a model, how to interact with the computer when formulating and solving a large model, how to efficiently perform what-if analysis with such a model, and so forth. Therefore, these details are deemphasized in this book. A student who becomes interested in conducting technical analyses as part of a management science team should plan to take additional, more technically oriented courses in management science. So what does an enlightened manager need to know about linear programming? A manager needs to have a good intuitive feeling for what linear programming is. One objective of this chapter is to begin to develop that intuition. That’s the purpose of studying the graphical method for solving two-variable problems. It is rare to have a real linear programming problem with as few as two decision variables. Therefore, the graphical method has essentially no practical value for solving real problems. However, it has great value for conveying the basic notion that linear programming involves pushing up against constraint boundaries and moving objective function values in a favorable direction as far as possible. Chapter 14, available at www.mhhe.com/Hillier6e, also demonstrates that this approach provides considerable geometric insight into how to analyze larger models by other methods. A manager must also have an appreciation for the relevance and power of linear programming to encourage its use where appropriate. For future managers using this book, this appreciation is being promoted by using application vignettes and their linked articles to describe real applications of linear programming and the resulting impact, as well as by including (in miniature form) various realistic examples and case studies that illustrate what can be done. Certainly a manager must be able to recognize situations where linear programming is applicable. We focus on developing this skill in Chapter 3, where you will learn how to recognize the identifying features for each of the major types of linear programming problems (and their mixtures). In addition, a manager should recognize situations where linear programming should not be applied. Chapter 8 will help to develop this skill by examining certain underlying assumptions of linear programming and the circumstances that violate these assumptions. That chapter also describes other approaches that can be applied where linear programming should not. A manager needs to be able to distinguish between competent and shoddy studies using linear programming (or any other management science technique). Therefore, another goal of the upcoming chapters is to demystify the overall process involved in conducting a management science study, all the way from first studying a problem to final implementation of the managerial decisions based on the study. This is one purpose of the case studies throughout the book. Finally, a manager must understand how to interpret the results of a linear programming study. He or she especially needs to understand what kinds of information can be obtained through what-if analysis, as well as the implications of such information for managerial decision making. Chapter 5 focuses on these issues. Review Questions 1. Does management generally get heavily involved with the technical details of a linear programming study? 2. What is the purpose of studying the graphical method for solving problems with two decision variables when essentially all real linear programming problems have more than two? 3. List the things that an enlightened manager should know about linear programming. 54 Chapter Two Linear Programming: Basic Concepts 2.9 Summary Linear programming is a powerful technique for aiding managerial decision making for certain kinds of problems. The basic approach is to formulate a mathematical model called a linear programming model to represent the problem and then to analyze this model. Any linear programming model includes decision variables to represent the decisions to be made, constraints to represent the restrictions on the feasible values of these decision variables, and an objective function that expresses the overall measure of performance for the problem. Spreadsheets provide a flexible and intuitive way of formulating and solving a linear programming model. The data are entered into data cells. Changing cells display the values of the decision variables, and an objective cell shows the value of the objective function. Output cells are used to help specify the constraints. After formulating the model on the spreadsheet, Solver is used to quickly find an optimal solution. Analytic Solver also provides a more powerful method for finding an optimal solution. The graphical method can be used to solve a linear programming model having just two decision variables. This method provides considerable insight into the nature of linear programming models and optimal solutions. Glossary absolute reference A reference to a cell (or a column or a row) with a fixed address, as indicated either by using a range name or by placing a $ sign in front of the letter and number of the cell reference. (Section 2.2 and Appendix A), 30 changing cells The cells in the spreadsheet that show the values of the decision variables. (Section 2.2), 28 constraint A restriction on the feasible values of the decision variables. (Sections 2.2 and 2.3), 32 constraint boundary equation The equation for the constraint boundary line. (Section 2.4), 37 constraint boundary line For linear programming problems with two decision variables, the line forming the boundary of the solutions that are permitted by the constraint. (Section 2.4), 37 data cells The cells in the spreadsheet that show the data of the problem. (Section 2.2), 28 decision variable An algebraic variable that represents a decision regarding the level of a particular activity. The value of the decision variable appears in a changing cell on the spreadsheet. (Section 2.3), 34 feasible region The geometric region that consists of all the feasible solutions. (Section 2.4), 35 feasible solution A solution that simultaneously satisfies all the constraints in the linear programming model. (Section 2.3), 34 functional constraint A constraint with a function of the decision variables on the left-hand side. All constraints in a linear programming model that are not nonnegativity constraints are called functional constraints. (Section 2.3), 34 graphical method A method for solving linear programming problems with two decision variables on a two-dimensional graph. (Section 2.4), 35 infeasible solution A solution that violates at least one of the constraints in the linear programming model. (Section 2.3), 34 linear programming model The mathematical model that represents a linear programming problem. (Sections 2.2 and 2.3), 28 nonnegativity constraint A constraint that expresses the restriction that a particular decision variable must be nonnegative (greater than or equal to zero). (Section 2.3), 34 objective cell The cell in the spreadsheet that shows the overall measure of performance of the decisions. (Section 2.2), 30 objective function The part of a linear programming model that expresses what needs to be either maximized or minimized, depending on the objective for the problem. The value of the objective function appears in the objective cell on the spreadsheet. (Section 2.3), 34 objective function line For a linear programming problem with two decision variables, a line whose points all have the same value of the objective function. (Section 2.4), 38 optimal solution The best feasible solution according to the objective function. (Section 2.3), 34 output cells The cells in the spreadsheet that provide output that depends on the changing cells. These cells frequently are used to help specify constraints. (Section 2.2), 29 parameter The parameters of a linear programming model are the constants (coefficients or right-hand sides) in the functional constraints and the objective function. Each parameter represents a quantity (e.g., the amount available of a resource) that is of importance for the analysis of the problem. (Section 2.3), 34 product-mix problem A type of linear programming problem where the objective is to find the most profitable mix of production levels for the products under consideration. (Section 2.1), 27 range name A descriptive name given to a cell or range of cells that immediately identifies what is there. (Section 2.2), 28 Chapter 2 Problems 55 relative reference A reference to a cell whose address is based upon its position relative to the cell containing the formula. (Section 2.2 and Appendix A), 30 solution Any single assignment of values to the decision variables, regardless of whether the assignment is a good one or even a feasible one. (Section 2.3), 34 Solver The spreadsheet tool that is used to specify the model in the spreadsheet and then to obtain an optimal solution for that model. (Section 2.5), 39 Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Excel Add-in: Excel Files: Analytic Solver Wyndor Example Supplement to This Chapter: Profit & Gambit Example More About the Graphical Method for Linear Programming Solved Problems The solutions are available at www.mhhe.com/Hillier6e. 2.S1. Back Savers Production Problem Back Savers is a company that produces backpacks primarily for students. They are considering offering some combination of two different models—the Collegiate and the Mini. Both are made out of the same rip-resistant nylon fabric. Back Savers has a long-term contract with a supplier of the nylon and receives a 5,000-square-foot shipment of the material each week. Each ­Collegiate requires 3 square feet while each Mini requires 2 square feet. The sales forecasts indicate that at most 1,000 Collegiates and 1,200 Minis can be sold per week. Each Collegiate requires 45 minutes of labor to produce and generates a unit profit of $32. Each Mini requires 40 minutes of labor and generates a unit profit of $24. Back Savers has 35 laborers that each provides 40 hours of labor per week. Management wishes to know what quantity of each type of backpack to produce per week in order to maximize the total profit. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. c. Use the graphical method by hand to solve this model. 2.S2. Conducting a Marketing Survey The marketing group for a cell phone manufacturer plans to conduct a telephone survey to determine consumer attitudes toward a new cell phone that is currently under development. In order to have a sufficient sample size to conduct the analysis, they need to contact at least 100 young males (under age 40), 150 older males (over age 40), 120 young females (under age 40), and 200 older females (over age 40). It costs $1 to make a daytime phone call and $1.50 to make an evening phone call (because of higher labor costs). This cost is incurred whether or not anyone answers the phone. The table below shows the likelihood of a given customer type answering each phone call. Assume the survey is conducted with whoever first answers the phone. Also, because of limited evening staffing, at most one-third of phone calls placed can be evening phone calls. How should the marketing group conduct the telephone survey so as to meet the sample size requirements at the lowest possible cost? a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. Who Answers? Young male Older male Young female Older female No answer Daytime Calls Evening Calls 10% 15% 20% 35% 20% 20% 30% 20% 25% 5% Problems We have inserted the symbol AS to the left of each problem or part where Analytic Solver is required. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 2.1. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 2.1. Briefly describe how linear programming was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 2.2. Reconsider the Wyndor Glass Co. case study introduced in Section 2.1. Suppose that the estimates of the unit profits for the two new products now have been revised to $600 for the doors and $300 for the windows. a. Formulate and solve the revised linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. c. Use the graphical method to solve this revised model. 56 Chapter Two Linear Programming: Basic Concepts 2.3. Reconsider the Wyndor Glass Co. case study introduced in Section 2.1. Suppose that Bill Tasto (Wyndor’s vice president for manufacturing) now has found a way to provide a little additional production time in Plant 2 to the new products. a. Use the graphical method to find the new optimal solution and the resulting total profit if one additional hour per week is provided. b. Repeat part a if two additional hours per week are provided instead. c. Repeat part a if three additional hours per week are provided instead. d. Use these results to determine how much each additional hour per week would be worth in terms of increasing the total profit from the two new products. 2.4. Use Solver to do Problem 2.3. 2.5. The following table summarizes the key facts about two products, A and B, and the resources, Q, R, and S, required to produce them. Resource Usage per Unit Produced Product A Product B Amount of Resource Available Q R S 2 1 3 1 2 3 2 2 4 Profit/unit $3,000 $2,000 Resource All the assumptions of linear programming hold. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. 2.6.* This is your lucky day. You have just won a $20,000 prize. You are setting aside $8,000 for taxes and partying expenses, but you have decided to invest the other $12,000. Upon hearing this news, two different friends have offered you an opportunity to become a partner in two different entrepreneurial ventures, one planned by each friend. In both cases, this investment would involve expending some of your time next summer as well as putting up cash. Becoming a full partner in the first friend’s venture would require an investment of $10,000 and 400 hours, and your estimated profit (ignoring the value of your time) would be $9,000. The corresponding figures for the second friend’s venture are $8,000 and 500 hours, with an estimated profit to you of $9,000. However, both friends are flexible and would allow you to come in at any fraction of a full partnership you would like. If you choose a fraction of a full partnership, all the above figures given for a full partnership (money investment, time investment, and your profit) would be multiplied by this same fraction. Because you were looking for an interesting summer job anyway (maximum of 600 hours), you have decided to participate in one or both friends’ ventures in whichever combination would maximize your total estimated profit. You now need to solve the problem of finding the best combination. a. Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Section 2.1. Then construct and fill in a table like Table 2.1 for this problem, identifying both the activities and the resources. b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve this model. e. Indicate why this spreadsheet model is a linear programming model. f. Formulate this same model algebraically. g. Identify the decision variables, objective function, nonnegativity constraints, functional constraints, and parameters in both the algebraic version and spreadsheet version of the model. h. Use the graphical method by hand to solve this model. What is your total estimated profit? 2.7. You are given the following linear programming model in algebraic form, where x1 and x2 are the decision variables and Z is the value of the overall measure of performance. Maximize Z = x 1 + 2 x 2 subject to Constraint on resource 1: x1 + x2 ≤ 5 (amount available) Constraint on resource 2: x1 + 3x2 ≤ 9 (amount available) and x 1 ≥ 0 x 2 ≥ 0 a. Identify the objective function, the functional constraints, and the nonnegativity constraints in this model. b. Incorporate this model into a spreadsheet. c. Is (x1, x2) = (3, 1) a feasible solution? d. Is (x1, x2) = (1, 3) a feasible solution? e. Use Solver to solve this model. 2.8. You are given the following linear programming model in algebraic form, where x1 and x2 are the decision variables and Z is the value of the overall measure of performance. Maximize Z = 3x 1 + 2x 2 subject to Constraint on resource 1: 3x1 + x2 ≤ 9 (amount available) Constraint on resource 2: x1 + 2x2 ≤ 8 (amount available) and x 1 ≥ 0 x 2 ≥ 0 a. Identify the objective function, the functional constraints, and the nonnegativity constraints in this model. b. Incorporate this model into a spreadsheet. c. Is (x1, x2) = (2, 1) a feasible solution? Chapter 2 Problems 57 d. Is (x1, x2) = (2, 3) a feasible solution? e. Is (x1, x2) = (0, 5) a feasible solution? f. Use Solver to solve this model. 2.9. The Whitt Window Company is a company with only three employees that makes two different kinds of handcrafted windows: a wood-framed and an aluminum framed window. They earn $60 profit for each wood-framed window and $30 profit for each aluminum-framed window. Doug makes the wood frames and can make 6 per day. Linda makes the aluminum frames and can make 4 per day. Bob forms and cuts the glass and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass. The company wishes to determine how many windows of each type to produce per day to maximize total profit. a. Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Section 2.1. Then construct and fill in a table like Table 2.1 for this problem, identifying both the activities and the resources. b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve this model. e. Indicate why this spreadsheet model is a linear programming model. f. Formulate this same model algebraically. g. Identify the decision variables, objective function, nonnegativity constraints, functional constraints, and parameters in both the algebraic version and spreadsheet version of the model. h. Use the graphical method to solve this model. i. A new competitor in town has started making wood-framed windows as well. This may force the company to lower the price it charges and so lower the profit made for each wood-framed window. How would the optimal solution change (if at all) if the profit per wood-framed window decreases from $60 to $40? From $60 to $20? j. Doug is considering lowering his working hours, which would decrease the number of wood frames he makes per day. How would the optimal solution change if he only makes 5 wood frames per day? 2.10. The Apex Television Company has to decide on the number of 65″ and 55″ sets to be produced at one of its factories. Market research indicates that at most 280 of the 65″ sets and 70 of the 55″ sets can be sold per month. The maximum number of work-hours available is 3,500 per month. A 65″ set requires 20 work-hours and a 55″ set requires 10 work-hours. Each 65″ set sold produces a profit of $270 and each 55″ set produces a profit of $180. A wholesaler has agreed to purchase all the television sets produced if the numbers do not exceed the maxima indicated by the market research. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. c. Use the graphical method to solve this model. 2.11. The WorldLight Company produces two light fixtures (Products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce per week so as to maximize profit. For each unit of Product 1, one unit of frame parts and two units of electrical components are required. For each unit of Product 2, three units of frame parts and two units of electrical components are required. The company has a weekly supply of 3,000 units of frame parts and 4,500 units of electrical components. Each unit of Product 1 gives a profit of $13, and each unit of Product 2, up to 900 units, gives a profit of $26. Any excess over 900 units of Product 2 brings no profit, so such an excess has been ruled out. a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. c. Formulate and solve a linear programming model for this problem on a spreadsheet. d. Formulate this same model algebraically. 2.12. The Primo Insurance Company is introducing two new product lines: special risk insurance and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per unit on mortgages. Management wishes to establish sales quotas for the new product lines to maximize total expected profit. The work requirements are shown below: Work-Hours per Unit Department Special Risk Mortgage Work-Hours Available Underwriting Administration Claims 3 0 2 2 1 0 2,400 800 1,200 a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. c. Formulate and solve a linear programming model for this problem on a spreadsheet. d. Formulate this same model algebraically. 2.13.* You are given the following linear programming model in algebraic form, with x1 and x2 as the decision variables and constraints on the usage of four resources: Maximize Profit = 2x 1 + x 2 58 Chapter Two Linear Programming: Basic Concepts subject to and x 2 ≤ 10 (resource 1) 2x 1 + 5x 2 ≤ 60 (resource 2) x 1 + x 2 ≤ 18 (resource 3) 3x 1 + x 2 ≤ 44 (resource 4) x 1 ≥ 0 x 2 ≥ 0 a. Use the graphical method to solve this model. b. Incorporate this model into a spreadsheet and then use Solver to solve this model. AS 2.14. Use Analytic Solver to formulate and solve the model shown in Problem 2.13 in a spreadsheet. 2.15. Because of your knowledge of management science, your boss has asked you to analyze a product mix problem involving two products and two resources. The model is shown below in algebraic form, where x1 and x2 are the production rates for the two products and P is the total profit. Maximize subject to and P = 3x 1 + 2x 2 x 1 + x 2≤ 8 (resource 1) 2 x 1 + x 2 ≤ 10 (resource 2) x 1 ≥ 0 x 2 ≥ 0 a. Use the graphical method to solve this model. b. Incorporate this model into a spreadsheet and then use Solver to solve this model. AS 2.16. Use Analytic Solver to formulate and solve the model shown in Problem 2.15 in a spreadsheet. 2.17. Independently consider each of the following changes in the Wyndor problem. In each case, apply the graphical method by hand to this new version of the problem, describe your conclusion, and then explain how and why the nature of this conclusion is different from the original Wyndor problem. a. The unit profit for the windows now is $200. b. To justify introducing these two new products, Wyndor management now requires that the total number of doors and windows produced per week must be at least 10. c. The functional constraints for Plants 2 and 3 now have been inadvertently deleted from the model. 2.18. Weenies and Buns is a food processing plant that manufactures buns and frankfurters for hot dogs. They grind their own flour for the buns at a maximum rate of 200 pounds per week. Each bun requires 0.1 pound of flour. They currently have a contract Item Beef tips Gravy Peas Carrots Dinner roll with Pigland, Inc., which specifies that a delivery of 800 pounds of pork product is delivered every Monday. Each frankfurter requires 1/4 pound of pork product. All the other ingredients in the buns and frankfurters are in plentiful supply. Finally, the labor force at Weenies and Buns consists of five employees working full time (40 hours per week each). Each bun requires two minutes of labor, and each frankfurter requires three minutes of labor. Each bun yields a profit of $0.20, and each frankfurter yields a profit of $0.40. Weenies and Buns would like to know how many buns and how many frankfurters they should produce each week so as to achieve the highest possible profit. a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. c. Formulate and solve a linear programming model for this problem on a spreadsheet. d. Formulate this same model algebraically. e. Use the graphical method to solve this model. 2.19. The Oak Works is a family-owned business that makes handcrafted dining room tables and chairs. They obtain the oak from a local tree farm, which ships them 2,500 pounds of oak each month. Each table uses 50 pounds of oak while each chair uses 25 pounds of oak. The family builds all the furniture itself and has 480 hours of labor available each month. Each table or chair requires six hours of labor. Each table nets Oak Works $400 in profit, while each chair nets $100 in profit. Since chairs are often sold with the tables, they want to produce at least twice as many chairs as tables. The Oak Works would like to decide how many tables and chairs to produce so as to maximize profit. AS a. Formulate and solve a linear programming model for this problem on a spreadsheet by using the Excel Solver. b. Use Analytic Solver to formulate and solve this model in a spreadsheet. c. Formulate this same model algebraically. 2.20. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 2.6. Briefly describe how linear programming was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 2.21. Nutri-Jenny is a weight-management center. It produces a wide variety of frozen entrées for consumption by its clients. The entrées are strictly monitored for nutritional content to ensure that the clients are eating a balanced diet. One new Calories (per oz.) Calories from Fat (per oz.) Vitamin A (IU per oz.) Vitamin C (mg per oz.) Protein (g, per oz.) Cost (per oz.) 54 20 15 8 40 19 15 0 0 10 0 0 15 350 0 0 1 3 1 0 8 0 1 1 1 40¢ 35¢ 15¢ 18¢ 10¢ Chapter 2 Problems 59 entrée will be a beef sirloin tips dinner. It will consist of beef tips and gravy, plus some combination of peas, carrots, and a dinner roll. Nutri-Jenny would like to determine what quantity of each item to include in the entrée to meet the nutritional requirements, while costing as little as possible. The nutritional information for each item and its cost are given in the table on the bottom of the preceding page. The nutritional requirements for the entrée are as follows: (1) it must have between 280 and 320 calories, (2) calories from fat should be no more than 30 percent of the total number of calories, and (3) it must have at least 600 IUs of vitamin A, 10 milligrams of vitamin C, and 30 grams of protein. Furthermore, for practical reasons, it must include at least 2 ounces of beef, and it must have at least half an ounce of gravy per ounce of beef. a. Formulate and solve a linear programming model for this problem on a spreadsheet by using the Excel Solver. AS b. Use Analytic Solver to formulate and solve this model in a spreadsheet. c. Formulate this same model algebraically. 2.22. Ralph Edmund loves steaks and potatoes. Therefore, he has decided to go on a steady diet of only these two foods (plus some liquids and vitamin supplements) for all his meals. Ralph realizes that this isn’t the healthiest diet, so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional requirements. He has obtained the following nutritional and cost information. Grams of Ingredient per Serving Steak Potatoes Daily Requirement (grams) Carbohydrates Protein Fat 5 20 15 15 5 2 ≥ 50 ≥ 40 ≤60 Cost per serving $4 $2 Ingredient Ralph wishes to determine the number of daily servings (may be fractional) of steak and potatoes that will meet these requirements at a minimum cost. a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. c. Formulate and solve a linear programming model for this problem on a spreadsheet. d. Formulate this same model algebraically. e. Use the graphical method by hand to solve this model. 2.23. Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an adequate supply of calories and vitamins. The cost, calorie content, and vitamin content of each feed is given in the table below. Contents Feed Type A Feed Type B Calories (per pound) Vitamins (per pound) Cost (per pound) 800 140 units $0.40 1,000 70 units $0.80 Each pig requires at least 8,000 calories per day and at least 700 units of vitamins. A further constraint is that no more than 1/3 of the diet (by weight) can consist of Feed Type A, since it contains an ingredient that is toxic if consumed in too large a quantity. a. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. b. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. c. Formulate and solve a linear programming model for this problem on a spreadsheet. d. Formulate this same model algebraically. 2.24. Reconsider the Profit & Gambit Co. problem described in Section 2.7. Suppose that the estimated data given in Table 2.2 now have been changed as shown in the table that accompanies this problem. a. Formulate and solve a linear programming model on a spreadsheet for this revised version of the problem. b. Formulate this same model algebraically. c. Use the graphical method to solve this model. d. What were the key changes in the data that caused your answer for the optimal solution to change from the one for the original version of the problem? Increase in Sales per Unit of Advertising Product Television Print Media Stain remover Liquid detergent Powder detergent Unit cost 0% 3 −1 $1 million 1.5% 4 2 $2 million Minimum Required Increase 3% 18 4 e. Write a paragraph to the management of the Profit & Gambit Co. presenting your conclusions from the above parts. Include the potential effect of further refining the key data in the above table. Also point out the leverage that your results might provide to management in negotiating a decrease in the unit cost for either of the advertising media. 2.25. You are given the following linear programming model in algebraic form, with x1 and x2 as the decision variables: Minimize Cost = 40x 1 + 50x 2 60 Chapter Two Linear Programming: Basic Concepts subject to Constraint 1: 2x 1 + 3x 2 ≥ 30 2: x 1 + x 2 ≥ 12 Constraint Constraint 3: 2x 1 + x 2 ≥ 20 and x 1 ≥ 0 x 2 ≥ 0 a. Use the graphical method to solve this model. b. How does the optimal solution change if the objective function is changed to Cost = 40x1 + 70x2? c. How does the optimal solution change if the third functional constraint is changed to 2x1 + x2 ≥ 15? d. Now incorporate the original model into a spreadsheet and use Solver to solve this model. e. Use Excel to do parts b and c. 2.26. The Learning Center runs a day camp for 6- to 10-yearolds during the summer. Its manager, Elizabeth Reed, is trying to reduce the center’s operating costs to avoid having to raise the tuition fee. Elizabeth is currently planning what to f eed Food Item Bread (1 slice) Peanut butter (1 tbsp) Jelly (1 tbsp) Apple Milk (1 cup) Cranberry juice (1 cup) the children for lunch. She would like to keep costs to a minimum, but also wants to make sure she is meeting the nutritional requirements of the children. She has already decided to go with peanut butter and jelly sandwiches, and some combination of apples, milk, and/or cranberry juice. The nutritional content of each food choice and its cost are given in the table that accompanies this problem. The nutritional requirements are as follows. Each child should receive between 300 and 500 calories, but no more than 30 percent of these calories should come from fat. Each child should receive at least 60 milligrams (mg) of vitamin C and at least 10 grams (g) of fiber. To ensure tasty sandwiches, Elizabeth wants each child to have a minimum of 2 slices of bread, 1 tablespoon (tbsp) of peanut butter, and 1 tbsp of jelly, along with at least 1 cup of liquid (milk and/or cranberry juice). Elizabeth would like to select the food choices that would minimize cost while meeting all these requirements. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Formulate this same model algebraically. Calories from Fat Total Calories Vitamin C (mg) Fiber (g) Cost (¢) 15 80 0 0 60 0 80 100 70 90 120 110 0 0 4 6 2 80 4 0 3 10 0 1 6 5 8 35 20 40 Case 2-1 Auto Assembly Automobile Alliance, a large automobile manufacturing company, organizes the vehicles it manufactures into three families: a family of trucks, a family of small cars, and a family of midsized and luxury cars. One plant outside Detroit, Michigan, assembles two models from the family of midsized and luxury cars. The first model, the Family Thrillseeker, is a four-door sedan with vinyl seats, plastic interior, standard features, and excellent gas mileage. It is marketed as a smart buy for middle-class families with tight budgets, and each Family Thrillseeker sold generates a modest profit of $3,600 for the company. The second model, the Classy Cruiser, is a two-door luxury sedan with leather seats, wooden interior, custom features, and navigational capabilities. It is marketed as a privilege of affluence for upper-middle-class families, and each Classy Cruiser sold generates a healthy profit of $5,400 for the company. Rachel Rosencrantz, the manager of the assembly plant, is currently deciding the production schedule for the next month. Specifically, she must decide how many Family Thrillseekers and how many Classy Cruisers to assemble in the plant to maximize profit for the company. She knows that the plant possesses a capacity of 48,000 labor-hours during the month. She also knows that it takes six labor-hours to assemble one Family Thrillseeker and 10.5 labor-hours to assemble one Classy Cruiser. Because the plant is simply an assembly plant, the parts required to assemble the two models are not produced at the plant. Instead, they are shipped from other plants around the Michigan area to the assembly plant. For example, tires, steering wheels, windows, seats, and doors all arrive from various supplier plants. For the next month, Rachel knows that she will only be able to obtain 20,000 doors from the door supplier. A recent labor strike forced the shutdown of that particular supplier plant for several days, and that plant will not be able to meet its production schedule for the next month. Both the Family Thrillseeker and the Classy Cruiser use the same door parts, with four needed for the Family Thrillseeker and two for the Classy Cruiser. In addition, a recent company forecast of the monthly demands for different automobile models suggests that the demand for the Classy Cruiser is limited to 3,500 cars. There is no limit on the demand for the Family Thrillseeker within the capacity limits of the assembly plant. Case 2-2 Cutting Cafeteria Costs 61 a. Formulate and solve a linear programming model to determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled. Before she makes her final production decisions, Rachel plans to explore the following questions independently, except where otherwise indicated. b. The marketing department knows that it can pursue a targeted $500,000 advertising campaign that will raise the demand for the Classy Cruiser next month by 20 percent. Should the campaign be undertaken? c. Rachel knows that she can increase next month’s plant capacity by using overtime labor. She can increase the plant’s labor-hour capacity by 25 percent. With the new assembly plant capacity, how many Family Thrillseekers and how many Classy Cruisers should be assembled? d. Rachel knows that overtime labor does not come without an extra cost. What is the maximum amount she should be willing to pay for all overtime labor beyond the cost of this labor at regular-time rates? Express your answer as a lump sum. e. Rachel explores the option of using both the targeted advertising campaign and the overtime labor-hours. The advertising campaign raises the demand for the Classy Cruiser by 20 percent, and the overtime labor increases the plant’s laborhour capacity by 25 percent. How many Family Thrillseekers and how many Classy Cruisers should be assembled using the advertising campaign and overtime labor-hours if the profit from each Classy Cruiser sold continues to be 50 ­percent more than for each Family Thrillseeker sold? f. Knowing that the advertising campaign costs $500,000 and the maximum usage of overtime labor-hours costs $1,600,000 beyond regular time rates, is the solution found in part e a wise decision compared to the solution found in part a? g. Automobile Alliance has determined that dealerships are actually heavily discounting the price of the Family Thrillseekers to move them off the lot. Because of a profit-­sharing agreement with its dealers, the company is not making a profit of $3,600 on the Family Thrillseeker but instead is making a profit of $2,800. Determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled given this new discounted profit. h. The company has discovered quality problems with the ­Family Thrillseeker by randomly testing Thrillseekers at the end of the assembly line. Inspectors have discovered that in over 60 percent of the cases, two of the four doors on a Thrillseeker do not seal properly. Because the percentage of defective Thrillseekers determined by the random testing is so high, the floor foreman has decided to perform quality control tests on every Thrillseeker at the end of the line. Because of the added tests, the time it takes to assemble one Family Thrillseeker has increased from 6 hours to 7.5 hours. Determine the number of units of each model that should be assembled given the new assembly time for the Family Thrillseeker. i. The board of directors of Automobile Alliance wishes to capture a larger share of the luxury sedan market and therefore would like to meet the full demand for Classy Cruisers. They ask Rachel to determine by how much the profit of her assembly plant would decrease as compared to the profit found in part a. They then ask her to meet the full demand for Classy Cruisers if the decrease in profit is not more than $2,000,000. j. Rachel now makes her final decision by combining all the new considerations described in parts f, g, and h. What are her final decisions on whether to undertake the advertising campaign, whether to use overtime labor, the number of Family Thrillseekers to assemble, and the number of Classy Cruisers to assemble? Case 2-2 Cutting Cafeteria Costs A cafeteria at All-State University has one special dish it serves like clockwork every Thursday at noon. This supposedly tasty dish is a casserole that contains sautéed onions, boiled sliced potatoes, green beans, and cream of mushroom soup. Unfortunately, students fail to see the special quality of this dish, and they loathingly refer to it as the Killer Casserole. The students reluctantly eat the casserole, however, because the cafeteria provides only a limited selection of dishes for Thursday’s lunch (namely, the casserole). Maria Gonzalez, the cafeteria manager, is looking to cut costs for the coming year, and she believes that one sure way to cut costs is to buy less expensive and perhaps lower quality ingredients. Because the casserole is a weekly staple of the cafeteria menu, she concludes that if she can cut costs on the ingredients purchased for the casserole, she can significantly reduce overall cafeteria operating costs. She therefore decides to invest time in determining how to minimize the costs of the casserole while maintaining nutritional and taste requirements. Maria focuses on reducing the costs of the two main ingredients in the casserole, the potatoes and green beans. These two ingredients are responsible for the greatest costs, nutritional content, and taste of the dish. Maria buys the potatoes and green beans from a wholesaler each week. Potatoes cost $0.40 per pound (lb), and green beans cost $1.00 per lb. All-State University has established nutritional requirements that each main dish of the cafeteria must meet. Specifically, the dish must contain 180 grams (g) of protein, 80 milligrams (mg) of iron, and 1,050 mg of vitamin C. (There are 454 g in one lb and 1,000 mg in one g.) For simplicity when planning, Maria assumes that only the potatoes and green beans contribute to the nutritional content of the casserole. Because Maria works at a cutting-edge technological university, she has been searching for information on the Internet to find the nutritional content of potatoes and green beans. Her research yields the following nutritional information about the two ingredients. Protein Iron Vitamin C Potatoes Green Beans 1.5 g per 100 g 0.3 mg per 100 g 12 mg per 100 g 5.67 g per 10 ounces 3.402 mg per 10 ounces 28.35 mg per 10 ounces (There are 28.35 g in one ounce.) 62 Chapter Two Linear Programming: Basic Concepts Edson Branner, the cafeteria cook who is surprisingly concerned about taste, informs Maria that an edible casserole must contain at least a six-to-five ratio in the weight of potatoes to green beans. Given the number of students who eat in the cafeteria, Maria knows that she must purchase enough potatoes and green beans to prepare a minimum of 10 kilograms (kg) of casserole each week. (There are 1,000 g in one kg.) Again, for simplicity in planning, she assumes that only the potatoes and green beans determine the amount of casserole that can be prepared. Maria does not establish an upper limit on the amount of casserole to prepare since she knows all leftovers can be served for many days thereafter or can be used creatively in preparing other dishes. a. Determine the amount of potatoes and green beans Maria should purchase each week for the casserole to minimize the ingredient costs while meeting nutritional, taste, and demand requirements. Before she makes her final decision, Maria plans to explore the following questions independently, except where otherwise indicated. b. Maria is not very concerned about the taste of the casserole; she is only concerned about meeting nutritional requirements and cutting costs. She therefore forces Edson to change the recipe to allow only for at least a one-to-two ratio in the weight of potatoes to green beans. Given the new recipe, determine the amount of potatoes and green beans Maria should purchase each week. c. Maria decides to lower the iron requirement to 65 mg since she determines that the other ingredients, such as the onions and cream of mushroom soup, also provide iron. Determine the amount of potatoes and green beans Maria should purchase each week given this new iron requirement. d. Maria learns that the wholesaler has a surplus of green beans and is therefore selling the green beans for a lower price of $0.50 per lb. Using the same iron requirement from part c and the new price of green beans, determine the amount of potatoes and green beans Maria should purchase each week. e. Maria decides that she wants to purchase lima beans instead of green beans since lima beans are less expensive and provide a greater amount of protein and iron than green beans. Maria again wields her absolute power and forces Edson to change the recipe to include lima beans instead of green beans. Maria knows she can purchase lima beans for $0.60 per lb from the wholesaler. She also knows that lima beans contain 22.68 g of protein and 6.804 mg of iron per 10 ounces of lima beans and no vitamin C. Using the new cost and nutritional content of lima beans, determine the amount of potatoes and lima beans Maria should purchase each week to minimize the ingredient costs while meeting nutritional, taste, and demand requirements. The nutritional requirements include the reduced iron requirement from part c. f. Will Edson be happy with the solution in part e? Why or why not? g. An All-State student task force meets during Body Awareness Week and determines that All-State University’s nutritional requirements for iron are too lax and that those for vitamin C are too stringent. The task force urges the university to adopt a policy that requires each serving of an entrée to contain at least 120 mg of iron and at least 500 mg of vitamin C. Using potatoes and lima beans as the ingredients for the dish and using the new nutritional requirements, determine the amount of potatoes and lima beans Maria should purchase each week. Case 2-3 Staffing a Call Center California Children’s Hospital has been receiving numerous customer complaints because of its confusing, decentralized appointment and registration process. When customers want to make appointments or register child patients, they must contact the clinic or department they plan to visit. Several problems exist with this current strategy. Parents do not always know the most appropriate clinic or department they must visit to address their children’s ailments. They therefore spend a significant amount of time on the phone being transferred from clinic to clinic until they reach the most appropriate clinic for their needs. The hospital also does not publish the phone numbers of all clinics and departments, and parents must therefore invest a large amount of time in detective work to track down the correct phone number. Finally, the various clinics and departments do not communicate with each other. For example, when a doctor schedules a referral with a colleague located in another department or clinic, that department or clinic almost never receives word of the referral. The parent must contact the correct department or clinic and provide the needed referral information. In efforts to reengineer and improve its appointment and registration process, the children’s hospital has decided to centralize the process by establishing one call center devoted exclusively to appointments and registration. The hospital is currently in the middle of the planning stages for the call center. Lenny Davis, the hospital manager, plans to operate the call center from 7 AM to 9 PM during the weekdays. Several months ago, the hospital hired an ambitious management consulting firm, Creative Chaos Consultants, to forecast the number of calls the call center would receive each hour of the day. Since all appointment and registration-related calls would be received by the call center, the consultants decided that they could forecast the calls at the call center by totaling the number of appointment and registration-related calls received by all clinics and departments. The team members visited all the clinics and departments, where they diligently recorded every call relating to appointments and registration. They then totaled these calls and altered the totals to account for calls missed during data collection. They also altered totals to account for repeat calls that occurred when the same parent called the hospital many times Case 2-3 Additional Case 63 because of the confusion surrounding the decentralized process. Creative Chaos Consultants determined the average number of calls the call center should expect during each hour of a weekday. The following table provides the forecasts. Work Shift Average Number of Calls 7 AM to 9 AM 9 AM to 11 AM 11 AM to 1 PM 1 PM to 3 PM 3 PM to 5 PM 5 PM to 7 PM 7 PM to 9 PM 40 calls per hour 85 calls per hour 70 calls per hour 95 calls per hour 80 calls per hour 35 calls per hour 10 calls per hour After the consultants submitted these forecasts, Lenny became interested in the percentage of calls from Spanish speakers since the hospital services many Spanish patients. Lenny knows that he has to hire some operators who speak Spanish to handle these calls. The consultants performed further data collection and determined that, on average, 20 percent of the calls were from Spanish speakers. Given these call forecasts, Lenny must now decide how to staff the call center during each two-hour shift of a week-day. During the forecasting project, Creative Chaos Consultants closely observed the operators working at the individual clinics and departments and determined the number of calls operators process per hour. The consultants informed Lenny that an operator is able to process an average of six calls per hour. Lenny also knows that he has both full-time and part-time workers available to staff the call center. A full-time employee works eight hours per day, but because of paperwork that must also be completed, the employee spends only four hours per day on the phone. To balance the schedule, the employee alternates the two-hour shifts between answering phones and completing paperwork. Fulltime employees can start their day either by answering phones or by completing paperwork on the first shift. The full-time employees speak either Spanish or English, but none of them are bilingual. Both Spanish-speaking and English-speaking employees are paid $15 per hour for work before 5 PM and $18 per hour for work after 5 PM. The full-time employees can begin work at the beginning of the 7 AM to 9 AM shift, 9 AM to 11 AM shift, 11 AM to 1 PM shift, or 1 PM to 3 PM shift. The part-time employees work for four hours, only answer calls, and only speak ­English. They can start work at the beginning of the 3 PM to 5 PM shift or the 5 PM to 7 pm shift, and, like the full-time employees, they are paid $15 per hour for work before 5 PM and $18 per hour for work after 5 PM. For the following analysis, consider only the labor cost for the time employees spend answering phones. The cost for paperwork time is charged to other cost centers. a. How many Spanish-speaking operators and how many ­English-speaking operators does the hospital need to staff the call center during each two-hour shift of the day in order to answer all calls? Please provide an integer number since half a human operator makes no sense. b. Lenny needs to determine how many full-time employees who speak Spanish, full-time employees who speak English, and part-time employees he should hire to begin on each shift. Creative Chaos Consultants advises him that linear programming can be used to do this in such a way as to minimize operating costs while answering all calls. Formulate a linear programming model of this problem. c. Obtain an optimal solution for the linear programming model formulated in part b to guide Lenny’s decision. d. Because many full-time workers do not want to work late into the evening, Lenny can find only one qualified Englishspeaking operator willing to begin work at 1 PM. Given this new constraint, how many full-time English-speaking operators, full-time Spanish-speaking operators, and part-time operators should Lenny hire for each shift to minimize operating costs while answering all calls? e. Lenny now has decided to investigate the option of hiring bilingual operators instead of monolingual operators. If all the operators are bilingual, how many operators should be working during each two-hour shift to answer all phone calls? As in part a, please provide an integer answer. f. If all employees are bilingual, how many full-time and parttime employees should Lenny hire to begin on each shift to minimize operating costs while answering all calls? As in part b, formulate a linear programming model to guide ­Lenny’s decision. g. What is the maximum percentage increase in the hourly wage rate that Lenny can pay bilingual employees over monolingual employees without increasing the total operating costs? h. What other features of the call center should Lenny explore to improve service or minimize operating costs? Source: This case is based on an actual project completed by a team of master’s students in what is now the Department of Management Science and Engineering at Stanford University. Additional Case Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business website, cases. ivey.uwo.ca/cases, in the segment of the Case-Mate area designated for this book. Chapter Three Linear Programming: Formulation and Applications Learning Objectives After completing this chapter, you should be able to 1. Recognize various kinds of managerial problems to which linear programming can be applied. 2. Describe the five major categories of linear programming problems, including their identifying features. 3. Formulate a linear programming model from a description of a problem in any of these categories. 4. Describe the difference between resource constraints and benefit constraints, including the difference in how they arise. 5. Describe fixed-requirement constraints and where they arise. 6. Identify the kinds of Excel functions that linear programming spreadsheet models use for the output cells, including the objective cell. 7. Identify the four components of any linear programming model and the kind of spreadsheet cells used for each component. 8. Understand the flexibility that managers have in prescribing key considerations that can be incorporated into a linear programming model. Chapter 2 introduced the basic concepts of linear programming, but only considered two tiny examples (the Wyndor case study introduced in Section 2.1 and the Profit & Gambit problem considered in Section 2.7) that only begin to illustrate the great power and versatility of this technique. This chapter will expand considerably further on the great application potential of linear programming by describing and illustrating several different types of linear programming problems. The many examples also will be supplemented by five solved problems presented at the end of the chapter. Linear programming problems come in many guises. And their models take various forms. This diversity can be confusing to both students and managers, making it difficult to recognize when linear programming can be applied to address a managerial problem. Since managers instigate management science studies, the ability to recognize the applicability of linear programming is an important managerial skill. This chapter focuses largely on developing this skill. The usual textbook approach to trying to teach this skill is to present a series of diverse examples of linear programming applications. The weakness of this approach is that it emphasizes differences rather than the common threads between these applications. Our approach will be to emphasize these common threads—the identifying features—that tie together linear programming problems even when they arise in very different contexts. We will describe some broad categories of linear programming problems and the identifying features that characterize them. Then we will use diverse examples, but with the purpose of illustrating and emphasizing the common threads among them. 64 3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 65 We will focus on the following five key categories of linear programming problems: • • • • • resource-allocation problems cost-benefit-trade-off problems mixed problems transportation problems assignment problems In each case, an important identifying feature is the nature of the restrictions on what decisions can be made, and thus the nature of the resulting functional constraints in the linear programming model. For each category, you will see how the basic data for a problem lead directly to a linear programming model with a certain distinctive form. Thus, model formulation becomes a by-product of proper problem formulation. The chapter begins with a case study (Section 3.1) that will illustrate the systematic procedure for formulating and solving the model for any linear programming problem. Section 3.2 will point out why the initial version of the case study turns out to be one example of a resource-allocation problem. However, we then return to the case study in Section 3.4, where additional managerial considerations turn the problem into a mixed problem. Sections 3.2 to 3.6 focus on the five categories of linear programming problems in turn. (Section 3.4 also briefly introduces still another category, fixed-requirements problems, where this category provides the framework for the more specific categories considered in Sections 3.5 and 3.6.) Section 3.7 then takes a broader look at the formulation of linear programming models from a managerial perspective. That section (along with Section 3.4) highlights the importance of having the model accurately reflect the managerial view of the problem. These (and other) sections also describe the flexibility available to managers for having the model structured to best fit their view of the important considerations. 3.1 A CASE STUDY: THE SUPER GRAIN CORP. ADVERTISING-MIX PROBLEM Claire Syverson, vice president for marketing of the Super Grain Corporation, is facing a daunting challenge: how to break into an already overly crowded breakfast cereal market in a big way. Fortunately, the company’s new breakfast cereal—Crunchy Start—has a lot going for it: Great taste. Nutritious. Crunchy from start to finish. She can recite the litany in her sleep now. It has the makings of a winning promotional campaign. However, Claire knows that she has to avoid the mistakes she made in her last campaign for a breakfast cereal. That had been her first big assignment since she won this promotion, and what a disaster! She thought she had developed a really good campaign. But somehow it had failed to connect with the most crucial segments of the market—young children and parents of young children. She also has concluded that it was a mistake not to include cents-off coupons in the magazine and newspaper advertising. Oh well. Live and learn. But she had better get it right this time, especially after the big stumble last time. The company’s president, David Sloan, already has impressed on her how important the success of Crunchy Start is to the future of the company. She remembers exactly how David concluded the conversation. “The company’s shareholders are not happy. We need to get those earnings headed in the right direction again.” Claire had heard this tune before, but she saw in David’s eyes how deadly serious he was this time. Claire often uses spreadsheets to help organize her planning. Her management science course in business school impressed upon her how valuable spreadsheet modeling can be. She regrets that she did not rely more heavily on spreadsheet modeling for the last campaign. That was a mistake that she is determined not to repeat. Now it is time for Claire to carefully review and formulate the problem in preparation for formulating a spreadsheet model. The Problem Claire already has employed a leading advertising firm, Giacomi & Jackowitz, to help design a nationwide promotional campaign that will achieve the largest possible exposure for Crunchy 66 Chapter Three Linear Programming: Formulation and Applications TABLE 3.1 Costs Cost and Exposure Data for the Super Grain Corp. Advertising-Mix Problem Cost Category Each TV Commercial Each Magazine Ad Each Sunday Ad Ad budget Planning budget $300,000 90,000 $150,000 30,000 $100,000 40,000 Expected number of exposures 1,300,000 600,000 500,000 Start. Super Grain will pay this firm a fee based on services performed (not to exceed $1 million) and has allocated an additional $4 million for advertising expenses. Giacomi & Jackowitz has identified the three most effective advertising media for this product: Medium 1: Television commercials on Saturday morning programs for children. Medium 2: Advertisements in food and family-oriented magazines. Medium 3: Advertisements in Sunday supplements of major newspapers. The problem now is to determine which levels should be chosen for these advertising activities to obtain the most effective advertising mix. To determine the best mix of activity levels for this particular advertising problem, it is necessary (as always) to identify the overall measure of performance for the problem and then the contribution of each activity toward this measure. An ultimate goal for Super Grain is to maximize its profits, but it is difficult to make a direct connection between advertising exposure and profits. Therefore, as a rough surrogate for profit, Claire decides to use expected number of exposures as the overall measure of performance, where each viewing of an advertisement by some individual counts as one exposure. Giacomi & Jackowitz has made preliminary plans for advertisements in the three media. The firm also has estimated the expected number of exposures for each advertisement in each medium, as given in the bottom row of Table 3.1. The number of advertisements that can be run in the different media are restricted by both the advertising budget (a limit of $4 million) and the planning budget (a limit of $1 million for the fee to Giacomi & Jackowitz). Another restriction is that there are only five commercial spots available for running different commercials (one commercial per spot) on children’s television programs Saturday morning (medium 1) during the time of the promotional campaign. (The other two media have an ample number of spots available.) Consequently, the three resources for this problem are: Resource 1: Advertising budget ($4 million). Resource 2: Planning budget ($1 million). Resource 3: TV commercial spots available (5). Table 3.1 shows how much of the advertising budget and the planning budget would be used by each advertisement in the respective media. • The first row gives the cost per advertisement in each medium. • The second row shows Giacomi & Jackowitz’s estimates of its total cost (including overhead and profit) for designing and developing each advertisement for the respective media.1 (This cost represents the billable fee from Super Grain.) • The last row then gives the expected number of exposures per advertisement. Analysis of the Problem Claire decides to formulate and solve a linear programming model for this problem on a spreadsheet. The formulation procedure summarized at the end of Section 2.2 guides this process. Like any linear programming model, this model will have four components: 1. The data 2. The decisions 1 When presenting its estimates in this form, the firm is making two simplifying assumptions. One is that its cost for designing and developing each additional advertisement in a medium is roughly the same as for the first advertisement in that medium. The second is that its cost when working with one medium is unaffected by how much work it is doing (if any) with the other media. 3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 67 3. The constraints 4. The measure of performance The spreadsheet needs to be formatted to provide the following kinds of cells for these components: Data → data cells Decisions → changing cells Constraints → output cells Measure of performance → objective cell Four kinds of cells are needed for these four components of a spreadsheet model. Figure 3.1 shows the spreadsheet model formulated by Claire. Let us see how she did this, including why she structured the spreadsheet in this way, by considering each of the ­components of the model individually. FIGURE 3.1 The spreadsheet model for the Super Grain problem (Section 3.1), including the objective cell TotalExposures (H13) and the other output cells BudgetSpent (F8:F9), as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C13:E13) show the optimal solution obtained by Solver. A 1 B D E TV Spots Magazine Ads SS Ads 1,300 600 500 C F G H Super Grain Corp. Advertising-Mix Problem 2 3 4 Exposures per Ad 5 (thousands) 6 Cost per Ad ($thousands) 7 Budget Budget Spent Available 8 Ad Budget 300 150 100 4,000 ≤ 4,000 9 Planning Budget 90 30 40 1,000 ≤ 1,000 TV Spots Magazine Ads SS Ads (thousands) 0 20 10 17,000 10 11 12 13 Number of Ads 14 15 Total Exposures ≤ Max TV Spots Solver Parameters Set Objective Cell: TotalExposures To: Max By Changing Variable Cells: NumberOfAds Subject to the Constraints: BudgetSpent <= BudgetAvailable TVSpots <= MaxTVSpots Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 5 F 6 Budget 7 8 9 Spent =SUMPRODUCT(C8:E8,NumberOfAds) =SUMPRODUCT(C9:E9,NumberOfAds) H 11 Total Exposures 12 (thousands) 13 =SUMPRODUCT(ExposuresPerAd,NumberOfAds) Range Name Cells BudgetAvailable BudgetSpent CostPerAd ExposuresPerAd MaxTVSpots NumberOfAds TotalExposures TVSpots H8: H9 F8: F9 C8: E9 C4: E4 C15 C13: E13 H13 C13 68 Chapter Three Linear Programming: Formulation and Applications The Data It is best to begin structuring the spreadsheet by deciding where to locate the data cells. Most of these cells normally are placed in the upper left-hand portion of the spreadsheet. The structure of the rest of the model will then follow the structure of the data cells. When some or all of the data are being obtained from a well-constructed table, the clarity of the spreadsheet usually is enhanced by retaining the format of the table when inserting the data into the spreadsheet. Claire followed these guidelines when she chose cells C4:E4 and C8:E9 to be the data cells to hold the information in Table 3.1 when using units of thousands of dollars. These data cells have been given the range names: ExposuresPerAd (C4:E4), CostPerAd (C8:E9), BudgetAvailable (H8:H9), and MaxTVSpots (C15). The other relevant kind of data here is the information given earlier about the amounts available of the three resources for the problem (the advertising budget, the planning budget, and the commercial spots available). Claire chose the locations of the data cells holding this information (H8:H9 and C15) to fit with the constraints that will be inserted a little later. The Decisions The problem has been defined as determining the most effective advertising mix among the three media selected by Giacomi & Jackowitz. Therefore, there are three decisions: Decision 1: TV = Number of commercials for separate spots on television. Decision 2: M = Number of advertisements in magazines. Decision 3: SS = Number of advertisements in Sunday supplements. The decisions already made for the locations of the data cells have designated columns C, D, and E to be the columns for these media. Therefore, the changing cells to hold these three decisions have been placed in row 13 in these columns: T V → cell C13 M → cell D13 SS → cell E13 These changing cells are collectively referred to by the range name NumberOfAds (C13:E13). The Constraints These changing cells need to be nonnegative. In addition, constraints are needed for the three resources. The first two resources are the ad budget and planning budget. Data for these two budgets already have ben placed in data cells in rows 8 and 9, so the constraints involving these budgets need to appear in these same two rows. The amounts spent from these two budgets are shown in the range BudgetSpent (F8:F9) and the amounts available from these budgets are shown in the range of data cells BudgetAvailable (H8:H9). As suggested by the ≤ signs entered into column G, the corresponding constraints are Total spending on advertising ≤ 4,000 (Ad budget in $1,000s) Total cost of planning ≤ 1,000 (Planning budget in $1,000s) Using the data in columns C, D, and E for the resources, these totals are Total spending on advertising = 300TV + 150M + 100SS Total cost of planning = 90TV + 30M + 40SS Excel Tip: Range names may overlap. For instance, we have used ­NumberOfAds to refer to the whole range of changing cells, C13:E13, and TVSpots to refer to the single cell, C13. These sums of products on the right-hand side are entered into the output cells BudgetSpent (F8:F9) by using the SUMPRODUCT functions shown in the lower right-hand side of Figure 3.1. Although the ≤ signs entered in column G are only cosmetic (trial solutions still can be entered in the changing cells that violate these inequalities), they will serve as a reminder later to use these same ≤ signs when entering the constraints in Solver. The third resource is TV spots for different commercials. Five such spots are available for purchase. The number of spots used is one of the changing cells (C13). Since this cell will be used in a constraint, we assign the cell its own range name: TVSpots (C13). The maximum number of TV spots available is in the data cell MaxTVSpots (C15). Thus, the required constraint is TVSpots ≤ MaxTVSpots. 3.1 A Case Study: The Super Grain Corp. Advertising-Mix Problem 69 The Measure of Performance Claire Syverson is using expected number of exposures as the overall measure of performance, so let Exposure = Expected number of exposures (in thousands) from all the advertising The data cells ExposuresPerAd (C4:E4) provide the expected number of exposures (in ­thousands) per advertisement in the respective media and the changing cells NumberOfAds (C13:E13) give the number of each type of advertisement. Therefore, Exposure = 1,300TV + 600M + 500SS = SUMPRODUCT (ExposuresPerAd, NumberOfAds) is the formula that needs to be entered into the objective cell, TotalExposures (H13). Summary of the Formulation The above analysis of the four components of the model has formulated the following linear programming model (in algebraic form) on the spreadsheet: Maximize Exposure = 1,300TV + 600M + 500SS subject to Ad spending: 300 TV + 150 M + 100 SS ≤ 4,000 costs: Planning 90 TV + 30 M + 40 SS ≤ 1,000 Number of television spots: TV ≤ 5 and T V ≥ 0 M ≥ 0 SS ≥ 0 The difficult work of defining the problem and gathering all the relevant data, including the information in Table 3.1, leads directly to this formulation. Excel Tip: With Excel’s Solver, the Solver dialog box is used to tell Solver the location on the spreadsheet of several of the elements of the model: the changing cells, the objective cell, and the constraints. With Analytic Solver, the Decisions, Constraints, and Objective menu on the Analytic Solver ribbon are used along with the Model pane. Solving the Model To solve the spreadsheet model formulated above, some key information needs to be entered into Solver. The lower left-hand side of Figure 3.1 shows the needed entries: the objective cell (TotalExposures), the changing cells (NumberOfAds), the goal of maximizing the objective cell, and the constraints BudgetSpent ≤ BudgetAvailable and TVSpots ≤ MaxTVSpots. Two options are also specified at the bottom of the Solver Parameters box on the lower left-hand side of Figure 3.1. The changing cells need nonnegativity constraints because negative values of advertising are not possible. Choose the Simplex LP (Excel’s Solver) or Standard LP/Quadratic Engine (Analytic Solver) solving method, because this is a linear programming model. Running Solver then finds an optimal solution for the model and displays it in the changing cells. The optimal solution given in row 13 of the spreadsheet provides the following plan for the promotional campaign: Do not run any television commercials. Run 20 advertisements in magazines. Run 10 advertisements in Sunday supplements. Since TotalExposures (H13) gives the expected number of exposures in thousands, this plan would be expected to provide 17,000,000 exposures. Evaluation of the Adequacy of the Model When she chose to use a linear programming model to represent this advertising-mix problem, Claire recognized that this kind of model does not provide a perfect match to this problem. However, a mathematical model is intended to be only an approximate representation of the real problem. Approximations and simplifying assumptions generally are required to have a workable model. All that is really needed is that there be a reasonably high correlation between the prediction of the model and what would actually happen in the real problem. The t eam now needs to check whether this criterion is satisfied. 70 Chapter Three Linear Programming: Formulation and Applications Linear programming models allow fractional solutions. Linear programming models should use SUM or SUMPRODUCT functions for the output cells, including the objective cell. One assumption of linear programming is that fractional solutions are allowed. For the current problem, this means that a fractional number (e.g., 3½) of television commercials (or of ads in magazines or Sunday supplements) should be allowed. This is technically true, since a commercial can be aired for less than a normal run, or an ad can be run in just a fraction of the usual magazines or Sunday supplements. However, one defect of the model is that it assumes that Giacomi & Jackowitz’s cost for planning and developing a commercial or ad that receives only a fraction of its usual run is only that fraction of its usual cost, even though the actual cost would be the same as for a full run. Fortunately, the optimal solution obtained was an integer solution (0 television commercials, 20 ads in magazines, and 10 ads in Sunday supplements), so the assumption that fractional solutions are allowed was not even needed. Although it is possible to have a fractional number of a normal run of commercials or ads, a normal run tends to be much more effective than a fractional run. Therefore, it would have been reasonable for Claire to drop the assumption that fractional solutions are allowed. If Claire had done this and the optimal solution for the linear programming model had not turned out to be integer, constraints can be added to require the changing cells to be integer. (The TBA Airlines example in the next section provides an illustration of this type of constraint.) After adding such constraints, the model is called an integer programming model instead of a linear programming model, but it still can be readily solved by Solver. Another key assumption of linear programming is that the appropriate equation for each of the output cells, including the objective cell, is one that can be expressed as a SUMPRODUCT of data cells and changing cells (or occasionally just a SUM of changing cells). For the objective cell (cell H13) in Figure 3.1, this implies that the expected number of exposures to be obtained from each advertising medium is proportional to the number of advertisements in that medium. This proportionality seems true, since each viewing of the advertisements by some individual counts as another exposure. Another implication of using a SUMPRODUCT function is that the expected number of exposures to be obtained from an advertising medium is unaffected by the number of advertisements in the other media. Again, this implication seems valid, since viewings of advertisements in different media count as separate exposures. Although a SUMPRODUCT function is appropriate for calculating the expected number of exposures, the choice of this number for the overall measure of performance is somewhat questionable. Management’s real objective is to maximize the profit generated as a result of the advertising campaign, but this is difficult to measure so expected number of exposures was selected to be a surrogate for profit. This would be valid if profit were proportional to the expected number of exposures. However, proportionality is only an approximation in this case because too many exposures for the same individual reach a saturation level where the impact (potential profit) from one more exposure is substantially less than for the first exposure. (When proportionality is not a reasonable approximation, Chapter 8 will describe nonlinear models that can be used instead.) To check how reasonable it is to use expected number of exposures as a surrogate for profit, Claire meets with Sid Jackowitz, one of the senior partners of Giacomi & Jackowitz. Sid indicates that the contemplated promotional campaign (20 advertisements in magazines and 10 in Sunday supplements) is a relatively modest one well below saturation levels. Most readers will only notice these ads once or twice, and a second notice is very helpful for reinforcing the first one. Furthermore, the readership of magazines and Sunday supplements is sufficiently different that the interaction of the advertising impact in these two media should be small. Consequently, Claire concludes that using expected number of exposures for the objective cell in Figure 3.1 provides a reasonable approximation. (A continuation of this case study in Case 8-1 will delve into the more complicated analysis that is required in order to use profit directly as the measure of performance to be recorded in the objective cell instead of making this approximation.) Next, Claire quizzes Sid about his firm’s costs for planning and developing advertisements in these media. Is it reasonable to assume that the cost in a given medium is proportional to the number of advertisements in that medium? Is it reasonable to assume that the cost of developing advertisements in one medium would not be substantially reduced if the firm had just finished developing advertisements in another medium that might have similar themes? Sid acknowledges that there is some carryover in ad planning from one medium to another, 3.2 Resource-Allocation Problems 71 especially if both are print media (e.g., magazines and Sunday supplements), but that the carryover is quite limited because of the distinct differences in these media. Furthermore, he feels that the proportionality assumption is quite reasonable for any given medium since the amount of work involved in planning and developing each additional advertisement in the medium is nearly the same as for the first one in the medium. The total fee that Super Grain will pay Giacomi & Jackowitz will eventually be based on a detailed accounting of the amount of work done by the firm. Nevertheless, Sid feels that the cost estimates previously provided by the firm (as entered in cells C9, D9, and E9 in units of thousands of dollars) give a reasonable basis for roughly projecting what the fee will be for any given plan (the entries in the changing cells) for the promotional campaign. Based on this information, Claire concludes that using a SUMPRODUCT function for cell F9 provides a reasonable approximation. Doing the same for cell F8 is clearly justified. Given her earlier conclusions as well, Claire decides that the linear programming model incorporated into Figure 3.1 (plus any expansions of the model needed later for the detailed planning) is a sufficiently accurate representation of the real advertising-mix problem. It will not be necessary to refine the results from this model by turning next to a more complicated kind of mathematical model (such as those to be described in Chapter 8). Therefore, Claire sends a memorandum to the company’s president, David Sloan, describing a promotional campaign that corresponds to the optimal solution from the linear programming model (no TV commercials, 20 ads in magazines, and 10 ads in Sunday supplements). She also requests a meeting to evaluate this plan and discuss whether some modifications should be made. We will pick up this story again in Section 3.4. Review Questions 3.2 1. What is the problem being addressed in this case study? 2. What overall measure of performance is being used? 3. What are the assumptions of linear programming that need to be checked to evaluate the adequacy of using a linear programming model to represent the problem under consideration? RESOURCE-ALLOCATION PROBLEMS In the opening paragraph of Chapter 2, we described managerial problems involving the allocation of an organization’s resources to its various productive activities. Those were resourceallocation problems. Resource-allocation problems are linear programming problems involving the allocation of resources to activities. The identifying feature for any such problem is that each functional constraint in the linear programming model is a resource constraint, which has the form Amount of resource used ≤ Amount of resource available for one of the resources. The amount of a resource used depends on which activities are undertaken, the levels of those activities, and how heavily those activities need to use the resource. Thus, the resource constraints place limits on the levels of the activities. The objective is to choose the levels of the activities so as to maximize some overall measure of performance (such as total profit) from the activities while satisfying all the resource constraints. Beginning with the Super Grain case study and then the Wyndor case study from ­Chapter 2, we will look at four examples that illustrate the characteristics of resource-allocation problems. These examples also demonstrate how this type of problem can arise in a variety of contexts. The Super Grain Corp. Advertising-Mix Problem The linear programming model formulated in Section 3.1 for the Super Grain case study is one example of a resource-allocation problem. The three activities under consideration are the advertising in the three types of media chosen by Giacomi & Jackowitz. 72 Chapter Three Linear Programming: Formulation and Applications Activity 1: TV commercials Activity 2: Magazine ads Activity 3: Sunday ads An initial step in formulating any resource-allocation problem is to identify the activities and the resources. The decisions being made are the levels of these activities, that is, the number of TV commercials, magazine ads, and Sunday ads to run. The resources to be allocated to these activities are Resource 1: Advertising budget ($4 million). Resource 2: Planning budget ($1 million). Resource 3: TV spots available for different commercials (5). where the amounts available of these resources are given in parentheses. Thus, this problem has three resource constraints: 1. Advertising budget used 2. Planning budget used 3. TV spots used ≤ $4 million ≤ $1 million ≤5 Rows 8–9 and cells C13:C15 in Figure 3.1 show these constraints in a spreadsheet. Cells C8:E9 give the amount of the advertising budget and the planning budget used by each unit of each activity, that is, the amount used by one TV spot, one magazine ad, and one Sunday ad, respectively. Cells C4, D4, and E4 on this spreadsheet give the contribution per unit of each activity to the overall measure of performance (expected number of exposures). Characteristics of Resource-Allocation Problems For each proposed activity, a decision needs to be made as to how much of the activity to do. In other words, what should the level of the activity be? These three kinds of data are needed for any resourceallocation problem. Other resource-allocation problems have the same kinds of characteristics as the Super Grain problem. In each case, there are activities where the decisions to be made are the levels of these activities. The contribution of each activity to the overall measure of performance is proportional to the level of that activity. Commonly, this measure of performance is the total profit from the activities, but occasionally it is something else (as in the Super Grain problem). Every problem of this type has a resource constraint for each resource. The amounts of the resources used depend on the levels of the activities. For each resource, the amount used by each activity is proportional to the level of that activity. The manager or management science team studying a resource allocation problem needs to gather (with considerable help) three kinds of data: 1. The amount available of each resource. 2. The amount of each resource needed by each activity. Specifically, for each combination of resource and activity, the amount of the resource used per unit of the activity must be estimated. 3. The contribution per unit of each activity to the overall measure of performance. Generally there is considerable work involved in developing these data. A substantial amount of digging and consultation is needed to obtain the best estimates available in a timely fashion. This step is critical. Well-informed estimates are needed to obtain a valid linear programming model for guiding managerial decisions. The dangers involved in inaccurate estimates are one reason why what-if analysis (Chapter 5) is such an important part of most linear programming studies. The Wyndor Glass Co. Product-Mix Problem The product-mix problem facing the management of the Wyndor Glass Co. in Section 2.1 is to determine the most profitable mix of production rates for the two new products, considering the limited availability of spare production capacity in the company’s three plants. This is a resource-allocation problem. The activities under consideration are Activity 1: Produce the special new doors. Activity 2: Produce the special new windows. 3.2 Resource-Allocation Problems 73 The decisions being made are the levels of these activities, that is, the production rates for the doors and windows. Production rate is being measured as the number of units (doors or windows) produced per week. Management’s objective is to maximize the total profit generated by the two new products, so the overall measure of performance is total profit. The contribution of each product to profit is proportional to the production rate for that product. The resources to be allocated to these activities are Resource 1: Production capacity in Plant 1. Resource 2: Production capacity in Plant 2. Resource 3: Production capacity in Plant 3. Each of the three functional constraints in the linear programming model formulated in Section 2.2 (see rows 7–9 of the spreadsheet in Figure 2.3 or 2.4) is a resource constraint for one of these three resources. Column E shows the amount of production capacity used in each plant and column G gives the amount available. Table 2.1 in Section 2.1 provides the data for the Wyndor problem. You already have seen how the numbers in Table 2.1 become the parameters in the linear programming model in either its spreadsheet formulation (Section 2.2) or its algebraic form (Section 2.3). The TBA Airlines Problem TBA Airlines is a small regional company that specializes in short flights in small passenger airplanes. The company has been doing well and management has decided to expand its operations. The Problem The basic issue facing management now is whether to purchase more small airplanes to add some new short flights or to start moving into the national market by purchasing some large airplanes for new cross-country flights (or both). Many factors will go into management’s final decision, but the most important one is which strategy is likely to be most profitable. The first row of Table 3.2 shows the estimated net annual profit (inclusive of capital recovery costs) from each type of airplane purchased. The second row gives the purchase cost per airplane and also notes that the total amount of capital available for airplane purchases is $250 million. The third row records the fact that management does not want to purchase more than five small airplanes because of limited possibilities for adding lucrative short flights, whereas they have not specified a maximum number for large airplanes (other than that imposed by the limited capital available). How many airplanes of each type should be purchased to maximize the total net annual profit? Formulation This is a resource-allocation problem. The activities under consideration are Activity 1: Purchase small airplanes. Activity 2: Purchase large airplanes. The decisions to be made are the levels of these activities, that is, S = Number of small airplanes to purchase L = Number of large airplanes to purchase The one resource to be allocated to these activities is Resource: Investment capital ($250 million). Thus, there is a single resource constraint: Investment capital spent ≤ $250 million TABLE 3.2 Data for the TBA Airlines Problem Small Airplane Net annual profit per airplane Purchase cost per airplane Maximum purchase quantity $7 million $25 million 5 Large Airplane $22 million $75 million No maximum Capital Available $250 million 74 Chapter Three Linear Programming: Formulation and Applications FIGURE 3.2 A spreadsheet model for the TBA Airlines integer programming problem where the changing cells, UnitsProduced (C12:D12), show the optimal airplane purchases obtained by Solver, and the objective cell, TotalProfit (G12), gives the resulting total profit in millions of dollars. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 B C D E F G <= Capital Available 250 TBA Airlines Airplane Purchasing Problem Unit Profit ($millions) Small Airplane 7 Capital per Unit Purchased 25 75 Capital ($millions) Number Purchased Maximum Small Airplanes 6 7 8 Small Airplane 1 <= 5 Capital Spent 250 Large Airplane 3 Total Profit ($millions) 73 E Capital Spent =SUMPRODUCT(CapitalPerUnitPurchased,NumberPurchased) Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: NumberPurchased Subject to the Constraints: CapitalSpent <= CapitalAvailable NumberPurchased = integer SmallAirplanes <= MaxSmallAirplanes Solver Options: Make Variables Nonnegative Solving Method: Simplex LP Large Airplane 22 10 11 12 G Total Profit ($millions) =SUMPRODUCT(UnitProfit,NumberPurchased) Range Name Capital Available CapitalPerUnitPurchased CapitalSpent MaxSmallAirplanes NumberPurchased SmallAirplanes TotalProfit UnitProfit Cells G8 C8:D8 E8 C14 C12:D12 C12 G12 C4:D4 In addition, management has specified one side constraint: Number of small airplanes purchased ≤ 5 Figure 3.2 shows the formulation of a spreadsheet model for this problem, where the data in Table 3.2 have been transferred into the data cells—UnitProfit (C4:D4), CapitalPerUnitPurchased (C8:D8), CapitalAvailable (G8), and MaxSmallAirplanes (C14). The resource constraint then appears in cells C8:G8 while C12:C14 shows the side constraint. The objective for this problem is to maximize the total net annual profit, so the equation for the objective cell is TotalProfit (G12) = SUMPRODUCT ( UnitProfit, UnitsPurchased) Since the TBA Airlines problem is a resource-allocation problem, this spreadsheet model has essentially the same form as the Super Grain and Wyndor problems except for one small difference. The changing cells in this case must have integer values since it is not feasible for the company to purchase and operate a fraction of an airplane. Therefore, constraints that the changing cells need to be integer are added. With Excel’s Solver, use the Add Constraint dialog box to choose the range of these cells (C12:D12) as the left-hand side and then choose int 3.2 Resource-Allocation Problems 75 Excel Tip: To constrain a range of changing cells to be integer in Excel’s Solver, choose the range of cells in the left-hand side of the Add Constraint dialog box and choose int from the pop-up menu. Clicking OK then enters the constraint that these cells = integer in the Solver dialog box. In Analytic Solver, select the range of cells to be constrained integer, and then under the Constraint menu on the Analytic Solver ribbon, choose Integer under the Variable Type/Bound submenu. Excel Tip: Even when a changing cell is constrained to be integer, rounding errors occasionally will cause Excel to return a noninteger value very close to an integer (e.g., 1.23E-10, meaning 0.000000000123). To make the spreadsheet cleaner, you may replace these “ugly” representations by their proper integer values in the changing cells. Excel Tip: In Excel’s Solver Options, the Integer Optimality (%) setting (1 percent by default) causes Solver to stop solving an integer programming problem when it finds a feasible solution whose objective function value is within the specified percentage of being optimal. In Analytic Solver, the equivalent setting is Integer Tolerance under the Engine tab of the Solver model (see Figure 2.19). This is useful to speed up solving large problems. For smaller problems (e.g., homework problems), this option should be set to 0 to guarantee finding an optimal solution. from the pop-up menu between the left-hand and right-hand side. In Analytic Solver, choose the changing cells (C12:D12), and then under the Constraint menu on the Analytic Solver ribbon, choose Integer under the Variable Type/Bound submenu. These changing cells in Figure 3.2 show the optimal solution, (S, L) = (1, 3), obtained after running Solver. One of the assumptions of linear programming is that the changing cells are allowed to have any values, including fractional values, that satisfy the functional and nonnegativity constraints. Therefore, technically speaking, the TBA problem is not a linear programming problem because of adding the constraints NumberPurchased = integer that are displayed in the Solver Parameters box in Figure 3.2. Such a problem that fits linear programming except for adding such constraints is called an integer programming problem. The method used by Solver to solve integer programming problems is quite different from that for solving linear programming problems. In fact, integer programming problems tend to be much more difficult to solve than linear programming problems so there is considerably more limitation on the size of the problem. However, this doesn’t matter to a spreadsheet modeler dealing with small problems. From his or her viewpoint, there is virtually no distinction between linear programming and integer programming problems. They are formulated in exactly the same way. Then, at the very end, a decision needs to be made as to whether any of the changing cells need to be restricted to integer values. If so, those constraints are added as described above. Keep this option in mind as we continue to discuss the formulation of various types of linear programming problems throughout the chapter. Summary of the Formulation The above formulation of a model with one resource constraint and one side constraint for the TBA Airlines problem now can be summarized (in algebraic form) as follows: Maximize subject to and Profit = 7S + 22L 25S + 75L ≤ 250 S ≤ 5 S ≥ 0 L ≥ 0 Capital Budgeting Financial planning is one of the most important areas of application for resource-allocation problems. The resources being allocated in this area are quite different from those for applications in the production planning area (such as the Wyndor Glass Co. product-mix problem), where the resources tend to be production facilities of various kinds. For financial planning, the resources tend to be financial assets such as cash, securities, accounts receivable, lines of credit, and so forth. Our specific example involves capital budgeting, where the resources are amounts of investment capital available at different points in time. The Problem The Think-Big Development Co. is a major investor in commercial real-estate development projects. It currently has the opportunity to share in three large construction projects: Project 1: Construct a high-rise office building. Project 2: Construct a hotel. Project 3: Construct a shopping center. Each project requires each partner to make investments at four different points in time: a down payment now, and additional capital after one, two, and three years. Table 3.3 shows for each project the total amount of investment capital required from all the partners at these four points in time. Thus, a partner taking a certain percentage share of a project is obligated to invest that percentage of each of the amounts shown in the table for the project. An Application Vignette A key part of a country’s financial infrastructure is its securities markets. By allowing a variety of financial institutions and their clients to trade stocks, bonds, and other financial securities, they help fund both public and private initiatives. Therefore, the efficient operation of its securities markets plays a crucial role in providing a platform for the economic growth of the country. Each central securities depository and its system for quickly settling security transactions are part of the operational backbone of securities markets and a key component of financial system stability. In Mexico, an institution called INDEVAL provides both the central securities depository and its security settlement system for the entire country. This security settlement system uses electronic book entries, modifying cash and securities balances, for the various parties in the transactions. The total value of the securities transactions the INDEVAL settles averages over $250 billion daily. This makes INDEVAL the main liquidity conduit for Mexico’s entire financial sector. Therefore, it is extremely important that INDEVAL’s system for clearing securities transactions be an exceptionally efficient one that maximizes the amount of cash that can be delivered almost instantaneously after the transactions. Because of past dissatisfaction with this system, INDEVAL’s board of directors ordered a major study in 2005 to completely redesign the system. Following more than 12,000 man-hours devoted to this redesign, the new system was successfully launched in N ­ ovember 2008. The core of the new system is a large linear programming model that is applied many times daily to choose which pending transactions should be settled immediately with the depositor’s available balances. Linear programming is ideally suited for this application because it can maximize the value of the transactions settled while taking into account the various relevant constraints. This application of linear programming has substantially enhanced and strengthened the Mexican financial infrastructure by reducing its daily liquidity requirements by $130 billion. It also reduces the intraday financing costs for market participants by more than $150 million annually. This application also led to INDEVAL winning the prestigious first prize in the 2010 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences. Source: D. Muñoz, M. de Lascurain, O. Romeo-Hernandez, F. Solis, L. de los Santoz, A. Palacios-Brun, F. Herrería, and J. Villaseñor, “INDEVAL Develops a New Operating and Settlement System Using Operations Research,” Interfaces 41, no. 1 (January–February 2011), pp. 8–17. (A link to this article is available at www.mhhe.com/Hillier6e.) TABLE 3.3 Financial Data for the Projects Being Considered for Partial Investment by the Think-Big Development Co. Investment Capital Requirements Year Office Building Hotel Shopping Center 0 1 2 3 $40 million 60 million 90 million 10 million $80 million 80 million 80 million 70 million $90 million 50 million 20 million 60 million $45 million $70 million $50 million Net present value 76 All three projects are expected to be very profitable in the long run. So the management of Think-Big wants to invest as much as possible in some or all of them. Management is willing to commit all the company’s investment capital currently available, as well as all additional investment capital expected to become available over the next three years. The objective is to determine the investment mix that will be most profitable, based on current estimates of profitability. Since it will be several years before each project begins to generate income, which will continue for many years thereafter, we need to take into account the time value of money in evaluating how profitable it might be. This is done by discounting future cash outflows (capital invested) and cash inflows (income), and then adding discounted net cash flows, to calculate a project’s net present value. Based on current estimates of future cash flows (not included here except for outflows), the estimated net present value for each project is shown in the bottom row of Table 3.3. All the investors, including Think-Big, then will split this net present value in proportion to their share of the total investment. For each project, participation shares are being sold to major investors, such as ThinkBig, who become the partners for the project by investing their proportional shares at the four specified points in time. For example, if Think-Big takes a 10 percent share of the office building, it will need to provide $4 million now, and then $6 million, $9 million, and $1 ­million in 1 year, 2 years, and 3 years, respectively. 3.2 Resource-Allocation Problems 77 The company currently has $25 million available for capital investment. Projections are that another $20 million will become available after one year, $20 million more after two years, and another $15 million after three years. What share should Think-Big take in the respective projects to maximize the total net present value of these investments? Formulation This is a resource-allocation problem. The activities under consideration are Activity 1: Invest in the construction of an office building. Activity 2: Invest in the construction of a hotel. Activity 3: Invest in the construction of a shopping center. Thus, the decisions to be made are the levels of these activities, that is, what participation share to take in investing in each of these projects. A participation share can be expressed as either a fraction or a percentage of the entire project, so the entire project is considered to be one “unit” of that activity. The resources to be allocated to these activities are the funds available at the four investment points. Funds not used at one point are available at the next point. (For simplicity, we will ignore any interest earned on these funds.) Therefore, the resource constraint for each point must reflect the cumulative funds to that point. Resource 1: Total investment capital available now. Resource 2: Cumulative investment capital available by the end of one year. Resource 3: Cumulative investment capital available by the end of two years. Resource 4: Cumulative investment capital available by the end of three years. Since the amount of investment capital available is $25 million now, another $20 million in one year, another $20 million in two years, and another $15 million in three years, the amounts available of the resources are the following: Amount of resource 1 available = $25million Amount of resource 2 available = $(25 + 20) million = $45 million Amount of resource 3 available = $(25 + 20 + 20) million = $65 million Amount of resource 4 available = $(25 + 20 + 20 + 15) million = $80 million Table 3.4 shows all the data involving these resources. The rightmost column gives the amounts of resources available calculated above. The middle columns show the cumulative amounts of the investment capital requirements listed in Table 3.3. For example, in the Office Building column of Table 3.4, the second number ($100 million) is obtained by adding the first two numbers ($40 million and $60 million) in the Office Building column of Table 3.3. The Data As with any resource-allocation problem, three kinds of data need to be gathered. One is the amounts available of the resources, as given in the rightmost column of Table 3.4. A second is the amount of each resource needed by each project, which is given in the middle columns of this table. A third is the contribution of each project to the overall measure of performance (net present value), as given in the bottom row of Table 3.3. The first step in formulating the spreadsheet model is to enter these data into data cells in the spreadsheet. In Figure 3.3, the data cells (and their range names) are NetPresentValue TABLE 3.4 Resource Data for the Think-Big Development Co. Investment-Mix Problem Cumulative Investment Capital Required for an Entire Project Resource 1 (Now) 2 (End of year 1) 3 (End of year 2) 4 (End of year 3) Office Building Hotel Shopping Center Amount of Resource Available $ 40 million 100 million 190 million 200 million $ 80 million 160 million 240 million 310 million $ 90 million 140 million 160 million 220 million $25 million 45 million 65 million 80 million 78 Chapter Three Linear Programming: Formulation and Applications FIGURE 3.3 The spreadsheet model for the Think-Big problem, including the formulas for the objective cell TotalNPV (H16) and the other output cells CapitalSpent (F9:F12), as well as the specifications needed to set up Solver. The changing cells ParticipationShare (C16:E16) show the optimal solution obtained by Solver. A 1 B C D E F G H Think-Big Development Co. Capital Budgeting Program 2 3 Office 4 Building Hotel Center 45 70 50 5 Net Present Value 6 ($millions) Shopping Cumulative 7 Cumulative Capital Required ($millions) 8 Cumulative Capital Capital Spent Available 9 Now 40 80 90 25 ≤ 25 10 End of Year 1 100 160 140 44.76 ≤ 45 11 End of Year 2 190 240 160 60.58 ≤ 65 12 End of Year 3 200 310 220 80 ≤ 80 13 14 Office Shopping Total NPV 15 Building Hotel Center ($millions) 0.00% 16.50% 13.11% 18.11 16 Participation Share Solver Parameters Set Objective Cell: TotalNPV To: Max By Changing Variable Cells: ParticipationShare Subject to the Constraints: CapitalSpent <= CapitalAvailable Solver Options: Make Variables Nonnegative Solving Method: Simplex LP F Range Name Cells CapitalAvailable CapitalRequired CapitalSpent ParticipationShare NetPresentValue TotalNPV H9:H12 C9:E12 F9:F12 C16:E16 C5:E5 H16 6 Cumulative 7 Capital 8 9 10 =SUMPRODUCT(C10:E10,ParticipationShare) 11 12 =SUMPRODUCT(C11:E11,ParticipationShare) =SUMPRODUCT C12:E12,ParticipationShare) Spent =SUMPRODUCT(C9:E9,ParticipationShare) H 14 Total NPV 15 ($millions) 16 =SUMPRODUCT(NetPresentValue,ParticipationShare) (C5:E5), CapitalRequired (C9:E12), and CapitalAvailable (H9:H12). To save space on the spreadsheet, these numbers are entered in units of millions of dollars. The Decisions With three activities under consideration, there are three decisions to be made. Decision 1: Decision 2: Decision 3: OB = Participation share in the office building. H = Participation share in the hotel. SC = Participation share in the shopping center. For example, if Think-Big management were to decide to take a one-tenth participation share (i.e., a 10 percent participation share) in each of these projects, then OB = 0.1 = 10% H = 0.1 = 10% SC = 0.1 = 10% However, it may not be desirable to take the same participation share (expressed as either a fraction or a percentage) in each of the projects, so the idea is to choose the best combination 3.2 Resource-Allocation Problems 79 of values of OB, H, and SC. In Figure 3.3, the participation shares (expressed as percentages) have been placed in changing cells under the data cells (row 16) in the columns for the three projects, so OB → cell C16 H → D16 SC → cell E16 where these cells are collectively referred to by the range name ParticipationShare (C16:E16). The Constraints The numbers in these changing cells make sense only if they are nonnegative, so the Make Variables Nonnegative option will need to be selected in the Excel’s Solver dialog box (or equivalently in Analytic Solver, set the Assume Non-Negative option to True in the Engine tab of the Model pane). In addition, the four resources require resource constraints: Total invested now ≤ 25 (millions of dollars available) Total invested within 1 year ≤ 45 (millions of dollars available) Total invested within 2 years ≤ 65 (millions of dollars available) Total invested within 3 years ≤ 80 (millions of dollars available) The data in columns C, D, and E indicate that (in millions of dollars) Total invested now = 40 OB + 80 H + 90 SC Total invested within 1 year = 100 OB + 160 H + 140 SC Total invested within 2 years = 190 OB + 240 H + 160 SC Total invested within 3 years = 200 OB + 310 H + 220 SC These totals are calculated in the output cells CapitalSpent (F9:F12) using the SUMPRODUCT function, as shown below the spreadsheet in Figure 3.3. Finally, ≤ signs are entered into column G to indicate the resource constraints that will need to be entered in Solver. The Measure of Performance The objective is to Maximize NPV = total net present value of the investments NetPresentValue (C5:E5) shows the net present value of each entire project, while ­ParticipationShare (C16:E16) shows the participation share for each of the projects. Therefore, the total net present value of all the participation shares purchased in all three projects is (in millions of dollars) NPV = 45 OB + 70 H + 50 SC ParticipationShare) = SUMPRODUCT ( NetPresentValue, → cell H16 Summary of the Formulation This completes the formulation of the linear programming model on the spreadsheet, as summarized below (in algebraic form). Maximize NPV = 45 OB + 70 H + 50 SC subject to Total invested now: 40 OB + 80 H + 90 SC ≤ 25 Total invested within 1 year: 100 OB + 160 H + 140 SC ≤ 45 Total invested within 2 years: 190 OB + 240 H + 160 SC ≤ 65 Total invested within 3 years: 200 OB + 310 H + 220 SC ≤ 80 and OB ≥ 0 H ≥ 0 SC ≥ 0 where all these numbers are in units of millions of dollars. 80 Chapter Three Linear Programming: Formulation and Applications Note that this model possesses the key identifying feature for resource-allocation problems, namely, each functional constraint is a resource constraint that has the form Amount of resource used ≤ Amount of resource available Solving the Model The lower left-hand side of Figure 3.3 shows the entries needed in Solver to specify the model, along with the selection of the usual two options. The spreadsheet shows the resulting optimal solution in row 16, namely, Invest nothing in the office building. Invest in 16.50 percent of the hotel. Invest in 13.11 percent of the shopping center. TotalNPV (H16) indicates that this investment program would provide a total net present value of $18.11 million. This amount actually is only an estimate of what the total net present value would turn out to be, depending on the accuracy of the financial data given in Table 3.3. There is some uncertainty about the construction costs for the three real estate projects, so the actual investment capital requirements for years 1, 2, and 3 may deviate somewhat from the amounts specified in this table. Because of the risk involved in these projects, the net present value for each one also might deviate from the amounts given at the bottom of the table. Chapter 5 describes one approach to analyzing the effect of such deviations. Chapters 12 and 13 will present another technique, called computer simulation, for systematically taking future uncertainties into account. Section 13.4 will focus on further analysis of this same example. Another Look at Resource Constraints These examples of resource-allocation problems illustrate a variety of resources: financial allocations for advertising and planning purposes, TV commercial spots available for purchase, available production capacities of different plants, the total amount of capital available for investment, and cumulative investment capital available by certain times. However, these illustrations only scratch the surface of the realm of possible resources that need to be allocated to activities in resource-allocation problems. In fact, by interpreting resource sufficiently broadly, any restriction on the decisions to be made that has the form Amount used ≤ Amount available can be thought of as a resource constraint, where the thing whose amount is being measured is the corresponding “resource.” Since any functional constraint with a ≤ sign in a linear programming model (including the side constraint limiting the number of small airplanes that can be purchased in the TBA Airlines example) can be verbalized in this form, any such constraint can be thought of as a resource constraint. Hereafter, we will use resource constraint to refer to any functional constraint with a ≤ sign in a linear programming model. The constant on the right-hand side represents the amount available of a resource. Therefore, the left-hand side represents the amount used of this resource. In the algebraic form of the constraint, the coefficient (positive or negative) of each decision variable is the resource usage per unit of the corresponding activity. Summary of the Formulation Procedure for Resource-Allocation Problems The four examples illustrate that the following steps are used for any resource-allocation problem to define the specific problem, gather the relevant data, and then formulate the linear programming model. 1. Since any linear programming problem involves finding the best mix of levels of various activities, identify these activities for the problem at hand. The decisions to be made are the levels of these activities. 2. From the viewpoint of management, identify an appropriate overall measure of performance (commonly profit, or a surrogate for profit) for solutions of the problem. 3.3 Cost–Benefit–Trade-Off Problems 81 FIGURE 3.4 A template of a spreadsheet model for pure resource-allocation problems. Activities Constraints Unit Profit Profit per unit of activity Resource used per unit of activity Resources Used SUMPRODUCT (resource used per unit, changing cells) Resources Available ≤ Total Profit Level of Activity Changing cells SUMPRODUCT(profit per unit, changing cells) 3. For each activity, estimate the contribution per unit of the activity to this overall measure of performance. 4. Identify the resources that must be allocated to the activities. 5. For each resource, identify the amount available and then the amount used per unit of each activity. 6. Enter the data gathered in steps 3 and 5 into data cells in a spreadsheet. A convenient format is to have the data associated with each activity in a separate column, the data for the unit profit and each constraint in a separate row, and to leave two blank columns between the activity columns and the amount of resource available column. Figure 3.4 shows a template of the overall format of a spreadsheet model for resource-allocation problems. 7. Designate changing cells for displaying the decisions on activity levels. 8. For the two blank columns created in step 6, use the left one as a Totals column for output cells and enter ≤ signs into the right one for all the resources. In the row for each resource, use the SUMPRODUCT function to enter the total amount used in the Totals column. 9. Designate an objective cell for displaying the overall measure of performance. Use a SUMPRODUCT function to enter this measure of performance. All the functional constraints in this linear programming model in a spreadsheet are resource constraints, that is, constraints with a ≤ sign. This is the identifying feature that classifies the problem as being a resource-allocation problem. Review Questions 3.3 1. 2. 3. 4. 5. What is the identifying feature for a resource-allocation problem? What is the form of a resource constraint? What are the three kinds of data that need to be gathered for a resource-allocation problem? Compare the types of activities for the four examples of resource-allocation problems. Compare the types of resources for the four examples of resource-allocation problems. COST–BENEFIT–TRADE-OFF PROBLEMS Cost–benefit–trade-off problems have a form that is very different from resource-allocation problems. The difference arises from managerial objectives that are very different for the two kinds of problems. For resource-allocation problems, limits are set on the use of various resources (including financial resources), and then the objective is to make the most effective use (according to some overall measure of performance) of these given resources. For cost–benefit–trade-off problems, management takes a more aggressive stance, prescribing what benefits must be achieved by the activities under consideration (regardless of 82 Chapter Three Linear Programming: Formulation and Applications A cost–benefit–trade-off formulation enables management to specify minimum goals for the benefits that need to be achieved by the activities. the resulting resource usage), and then the objective is to achieve all these benefits with minimum cost. By prescribing a minimum acceptable level for each kind of benefit, and then minimizing the cost needed to achieve these levels, management hopes to obtain an appropriate trade-off between cost and benefits. (You will see in Chapter 5 that what-if analysis plays a key role in providing the additional information needed for management to choose the best trade-off between cost and benefits.) Cost–benefit–trade-off problems are linear programming problems where the mix of levels of various activities is chosen to achieve minimum acceptable levels for various benefits at a minimum cost. The identifying feature is that each functional constraint is a benefit constraint, which has the form Level achieved ≥ Minimum acceptable level for one of the benefits. These three kinds of data are needed for any cost– benefit–trade-off problem. Interpreting benefit broadly, we can think of any functional constraint with a ≥ sign as a benefit constraint. In most cases, the minimum acceptable level will be prescribed by management as a policy decision, but occasionally this number will be dictated by other circumstances. For any cost–benefit–trade-off problem, a major part of the study involves identifying all the activities and benefits that should be considered and then gathering the data relevant to these activities and benefits. Three kinds of data are needed: 1. The minimum acceptable level for each benefit (a managerial policy decision). 2. For each benefit, the contribution of each activity to that benefit (per unit of the activity). 3. The cost per unit of each activity. Let’s examine two examples of cost–benefit–trade-off problems. The Profit & Gambit Co. Advertising-Mix Problem An initial step in formulating any cost–benefit–­ tradeoff problem is to identify the activities and the benefits. As described in Section 2.7, the Profit & Gambit Co. will be undertaking a major new advertising campaign focusing on three cleaning products. The two kinds of advertising to be used are television and the print media. Management has established minimum goals—the minimum acceptable increase in sales for each product—to be gained by the campaign. The problem is to determine how much to advertise in each medium to meet all the sales goals at a minimum total cost. The activities in this cost–benefit–trade-off problem are Activity 1: Advertise on television. Activity 2: Advertise in the print media. The benefits being sought from these activities are Benefit 1: Increased sales for a spray prewash stain remover. Benefit 2: Increased sales for a liquid laundry detergent. Benefit 3: Increased sales for a powder laundry detergent. Management wants these increased sales to be at least 3 percent, 18 percent, and 4 percent, respectively. As shown in Section 2.7, each benefit leads to a benefit constraint that incorporates the managerial goal for the minimum acceptable level of increase in the sales for the corresponding product, namely, Level of benefit 1 achieved ≥ 3% Level of benefit 2 achieved ≥ 18% Level of benefit 3 achieved ≥ 4% The data for this problem are given in Table 2.2 (Section 2.7). Section 2.7 describes how the linear programming model is formulated directly from the numbers in this table. This example provides an interesting contrast with the Super Grain Corp. case study in ­Section 3.1, which led to a formulation as a resource-allocation problem. Both are advertising-mix 3.3 Cost–Benefit–Trade-Off Problems 83 problems, yet they lead to entirely different linear programming models. They differ because of the differences in the managerial view of the key issues in each case: • As the vice president for marketing of Super Grain, Claire Syverson focused first on how much to spend on the advertising campaign and then set limits (an advertising budget of $4 million and a planning budget of $1 million) that led to resource constraints. • The management of Profit & Gambit instead focused on what it wanted the advertising campaign to accomplish and then set goals (minimum required increases in sales) that led to benefit constraints. From this comparison, we see that it is not the nature of the application that determines the classification of the resulting linear programming formulation. Rather, it is the nature of the restrictions imposed on the decisions regarding the mix of activity levels. If the restrictions involve limits on the usage of resources, that identifies a resource-allocation problem. If the restrictions involve goals on the levels of benefits, that characterizes a cost–benefit–trade-off problem. Frequently, the nature of the restrictions arise from the way management frames the problem. However, we don’t want you to get the idea that every linear programming problem falls entirely and neatly into either one type or the other. In the preceding section and this one, we are looking at pure resource-allocation problems and pure cost–benefit–trade-off problems. Although many real problems tend to be either one type or the other, it is fairly common to have both resource constraints and benefit constraints, even though one may predominate. (In the next section, you will see an example of how both types of constraints can arise in the same problem when the management of the Super Grain Corp. introduces additional considerations into the analysis of their advertising-mix problem.) Furthermore, we still need to consider additional categories of linear programming problems in the remaining sections of this chapter. Now, another example of a pure cost–benefit–trade-off problem. Personnel Scheduling One of the common applications of cost–benefit–trade-off analysis involves personnel scheduling for a company that provides some kind of service, where the objective is to schedule the work times of the company’s employees so as to minimize the cost of providing the level of service specified by management. The following example illustrates how this can be done. The Problem Union Airways is adding more flights to and from its hub airport and so needs to hire additional customer service agents. However, it is not clear just how many more should be hired. Management recognizes the need for cost control while also consistently providing a satisfactory level of service to the company’s customers, so a desirable trade-off between these two factors is being sought. Therefore, a management science team is studying how to schedule the agents to provide satisfactory service with the smallest personnel cost. Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service. (The queueing models presented in Chapter 11 can be used to determine the minimum numbers of agents needed to reduce the waiting times incurred by customers down to reasonable levels.) These numbers are shown in the last column of Table 3.5 for the time periods given in the first column. The other entries in this table reflect one of the provisions in the company’s current contract with the union that represents the customer service agents. The provision is that each agent works an eight-hour shift. The authorized shifts are Shift 1: 6:00 AM to 2:00 PM. Shift 2: 8:00 AM to 4:00 PM. Shift 3: Noon to 8:00 PM. Shift 4: 4:00 PM to midnight. Shift 5: 10:00 PM to 6:00 AM. 84 Chapter Three Linear Programming: Formulation and Applications TABLE 3.5 Data for the Union Airways Personnel Scheduling Problem Time Periods Covered by Shift Time Period 1 6:00 AM to 8:00 AM 8:00 AM to 10:00 AM 10:00 AM to noon Noon to 2:00 PM 2:00 PM to 4:00 PM 4:00 PM to 6:00 PM 6:00 PM to 8:00 PM 8:00 PM to 10:00 PM 10:00 PM to midnight Midnight to 6:00 AM ✓ ✓ ✓ ✓ Daily cost per agent $170 2 ✓ ✓ ✓ ✓ $160 3 ✓ ✓ ✓ ✓ $175 4 ✓ ✓ ✓ ✓ $180 5 Minimum Number of Agents Needed ✓ ✓ 48 79 65 87 64 73 82 43 52 15 $195 Check marks in the main body of Table 3.5 show the time periods covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the contract differ by shift. For each shift, the daily compensation (including benefits) for each agent is shown in the bottom row. The problem is to determine how many agents should be assigned to the respective shifts each day to minimize the total personnel cost for agents, based on this bottom row, while meeting (or surpassing) the service requirements given in the last column. Formulation This problem is, in fact, a pure cost–benefit–trade-off problem. To formulate the problem, we need to identify the activities and benefits involved. Activities correspond to shifts. The level of each activity is the number of agents assigned to that shift. A unit of each activity is one agent assigned to that shift. Thus, the general description of a linear programming problem as finding the best mix of activity levels can be expressed for this specific application as finding the best mix of shift sizes. Benefits correspond to time periods. For each time period, the benefit provided by the activities is the service that agents provide customers during that period. The level of a benefit is measured by the number of agents on duty during that time period. Once again, a careful formulation of the problem, including gathering all the relevant data, leads rather directly to a spreadsheet model. This model is shown in Figure 3.5, and we outline its formulation below. The Data As indicated in this figure, all the data in Table 3.5 have been entered directly into the data cells CostPerShift (C5:G5), ShiftWorksTimePeriod (C8:G17), and MinimumNeeded (J8:J17). For the ShiftWorksTimePeriod (C8:G17) data, an entry of 1 indicates that the corresponding shift includes that time period whereas 0 indicates not. Like any cost–­ benefit–tradeoff problem, these numbers indicate the contribution of each activity to each benefit. Each agent working a shift contributes either 0 or 1 toward the minimum number of agents needed in a time period. The Decisions Since the activities in this case correspond to the five shifts, the decisions to be made are S1 = Number of agents to assign to Shift 1 (starts at 6 AM) S2 = Number of agents to assign to Shift 2 (starts at 8 AM) 3.3 Cost–Benefit–Trade-Off Problems 85 FIGURE 3.5 The spreadsheet model for the Union Airways problem, including the formulas for the objective cell TotalCost (J21) and the other output cells TotalWorking (H8:H17), as well as the specifications needed to set up Solver. The changing cells NumberWorking (C21:G21) show the optimal solution obtained by Solver. A 1 B D C E F G H I J Union Airways Personnel Scheduling Problem 2 3 6AM–2PM 8AM–4PM Noon–8PM 4PM–Midnight 10PM–6AM 4 Shift Shift Shift Shift Shift $170 $160 $175 $180 $195 5 Cost per Shift 6 Total Minimum Working 7 Time Period 8 6AM–8AM 1 0 0 0 0 9 8AM–10AM 1 1 0 0 0 Shift Works Time Period? (1=yes, 0=no) Needed 48 ≥ 48 79 ≥ 79 10 10AM–12PM 1 1 0 0 0 79 ≥ 65 11 12PM–2PM 1 1 1 0 0 118 ≥ 87 12 2PM–4PM 0 1 1 0 0 70 ≥ 64 13 4PM–6PM 0 0 1 1 0 82 ≥ 73 14 6PM–8PM 0 0 1 1 0 82 ≥ 82 15 8PM–10PM 0 0 0 1 0 43 ≥ 43 16 10PM–12AM 0 0 0 1 1 58 ≥ 52 15 ≥ 15 17 12AM–6AM 0 0 0 0 1 6AM–2PM 8AM–4PM Noon–8PM 4PM–Midnight 10PM–6AM Shift Shift Shift Shift Shift Total Cost 48 31 39 43 15 $30,610 18 19 20 21 Number Working Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells: NumberWorking Subject to the Constraints: NumberWorking = integer TotalWorking >= MinimumNeeded Solver Options: Make Variables Nonnegative Solving Method: Simplex LP Range Name Cells CostPerShift MinimumNeeded NumberWorking ShiftWorksTimePeriod TotalCost TotalWorking C5:G5 J8:J17 C21:G21 C8:G17 J21 H8:H17 H 6 Total 7 Working 8 9 10 =SUMPRODUCT(C8:G8,NumberWorking) =SUMPRODUCT(C9:G9,NumberWorking) =SUMPRODUCT(C10:G10,NumberWorking) 11 =SUMPRODUCT(C11:G11,NumberWorking) 12 =SUMPRODUCT(C12:G12,NumberWorking) 13 14 15 16 17 =SUMPRODUCT(C13:G13,NumberWorking) =SUMPRODUCT(C14:G14,NumberWorking) =SUMPRODUCT(C15:G15,NumberWorking) =SUMPRODUCT(C16:G16,NumberWorking) =SUMPRODUCT(C17:G17,NumberWorking) J 20 Total Cost 21 =SUMPRODUCT(CostPerShift,NumberWorking) S3 = Number of agents to assign to Shift 3 (starts at noon) S4 = Number of agents to assign to Shift 4 (starts at 4 PM) S5 = Number of agents to assign to Shift 5 (starts at 10 PM) The changing cells to hold these numbers have been placed in the activity columns in row 21, so S 1 → cell C21 S 2 → cell D21 . . . S 5 → cell G21 86 Chapter Three Linear Programming: Formulation and Applications where these cells are collectively referred to by the range name NumberWorking (C21:G21). The Constraints These changing cells need to be nonnegative. In addition, we need 10 benefit constraints, where each one specifies that the total number of agents serving in the corresponding time period listed in column B must be no less than the minimum acceptable number given in column J. Thus, these constraints are Total number of agents serving 6 − 8 am ≥ 48 (min. acceptable) Total number of agents serving 8 − 10 am ≥ 79 (min. acceptable) . . . Total number of agents serving midnight − 6 am ≥ 15 (min. acceptable) Since columns C to G indicate which of the shifts serve each of the time periods, these totals are Total number of agents serving 6 − 8 am = S 1 Total number of agents serving 8 − 10 am = S 1 + S 2 . . . Total number of agents serving midnight − 6 am = S 5 These totals are calculated in the output cells TotalWorking (H8:H17) using the SUMPRODUCT functions shown below the spreadsheet in Figure 3.5. One other type of constraint is that the number of agents assigned to each shift must have an integer value. These constraints for the five shifts should be added in the same way as described for the TBA Airlines problem in Section 3.2. In particular, with Excel’s Solver they are added in the Add Constraint dialog box by entering NumberWorking on the left-hand side and then choosing int from the pop-up menu between the left-hand side and the right-hand side. The set of constraints, NumberWorking = integer, then appears in the Solver Parameters, as shown in Figure 3.5. In Analytic Solver, select the range of cells to be constrained integer, and then under the Constraint menu on the Analytic Solver ribbon, choose Integer under the Variable Type/Bound submenu. The Measure of Performance The objective is to Minimize Cost = Total daily personnel cost for all agents Since CostPerShift (C5:G5) gives the daily cost per agent on each shift and NumberWorking (C21:G21) gives the number of agents working each shift, Cost = 170 S 1 + 160 S 2 + 175 S 3 + 180 S 4 + 195 S 5 (in dollars) = SUMPRODUCT (CostPerShift, NumberWorking) → cell J21 Summary of the Formulation The above steps provide the complete formulation of the linear programming model on a spreadsheet, as summarized below (in algebraic form). Minimize Cost = 170 S 1 + 160 S 2 + 175 S 3 + 180 S 4 + 195 S 5 subject to Total agents 6 − 8 am S 1 ≥ 48 Total agents 8 − 10 am S 1 + S 2 ≥ 79 . . . Total agents midnight − 6 am S 5 ≥ 15 (in dollars) 3.4 Mixed Problems 87 FIGURE 3.6 A template of a spreadsheet model for pure cost–benefit–trade-off problems. Unit Cost Cost per unit of activity Constraints Activities Benefit achieved per unit of activity Benefit Achieved Benefit Needed SUMPRODUCT (benefit per unit, changing cells) ≥ Total Cost Level of Activity Changing cells SUMPRODUCT(cost per unit, changing cells) and S 1 ≥ 0 S 2 ≥ 0 S 3 ≥ 0 S 4 ≥ 0 S 5 ≥ 0 Solving the Model The lower left-hand corner of Figure 3.5 shows the entries needed in Solver, along with the selection of the usual two options. After solving, NumberWorking (C21:G21) in the spreadsheet shows the resulting optimal solution for the number of agents that should be assigned to each shift. TotalCost (J21) indicates that this plan would cost $30,610 per day. Summary of the Formulation Procedure for Cost–Benefit–Trade-Off Problems The nine steps in formulating any cost–benefit–trade-off problem follow the same pattern as presented at the end of the preceding section for resource-allocation problems, so we will not repeat them here. The main differences are that the overall measure of performance now is the total cost of the activities (or some surrogate of total cost chosen by management) in steps 2 and 3, benefits now replace resources in steps 4 and 5, and ≥ signs now are entered to the right of the output cells for benefits in step 8. Figure 3.6 shows a template of the format of a spreadsheet model for cost–benefit–trade-off problems. All the functional constraints in the resulting model are benefit constraints, that is, constraints with a ≥ sign. This is the identifying feature of a pure cost–benefit–trade-off problem. Review Questions 3.4 1. What is the difference in managerial objectives between resource-allocation problems and cost–benefit–trade-off problems? 2. What is the identifying feature of a cost–benefit–trade-off problem? 3. What is the form of a benefit constraint? 4. What are the three kinds of data that need to be gathered for a cost–benefit–trade-off problem? 5. Compare the types of activities for the two examples of cost–benefit–trade-off problems. 6. Compare the types of benefits for the two examples of cost–benefit–trade-off problems. MIXED PROBLEMS Sections 3.2 and 3.3 each described a broad category of linear programming problems— resource-allocation and cost–benefit–trade-off problems. As summarized in Table 3.6, each features one of the first two types of functional constraints shown there. In fact, the identifying feature of a pure resource-allocation problem is that all its functional constraints are resource constraints. The identifying feature of a pure cost–benefit–trade-off problem is that all its functional constraints are benefit constraints. (Keep in mind that the functional constraints include all the constraints of a problem except its nonnegativity constraints.) 88 Chapter Three Linear Programming: Formulation and Applications The bottom row of Table 3.6 shows the last of the three types of functional constraints, namely, fixed-requirement constraints, which require that the left-hand side of each such constraint must exactly equal some fixed amount. Thus, since the left-hand side represents the amount provided of some quantity, the form of a fixed-requirement constraint is Amount provided = Required amount The identifying feature of a pure fixed-requirements problem is that it is a linear programming problem where all its functional constraints are fixed-requirement constraints. The next two sections will describe two particularly prominent types of fixed-requirement problems called transportation problems and assignment problems. However, before turning to these types of problems, we first will use a continuation of the Super Grain case study from Section 3.1 to illustrate how many linear programming problems fall into another broad category called mixed problems. Many linear programming problems do not fit completely into any of the previously discussed categories (pure resource-allocation problems, cost–benefit–trade-off problems, and fixedrequirement problems) because the problem’s functional constraints include more than one of the types shown in Table 3.6. Such problems are called mixed problems. Now let us see how a more careful analysis of the Super Grain case study turns this resourceallocation problem into a mixed problem that includes all three types of functional constraints shown in Table 3.6. Super Grain Management Discusses Its Advertising-Mix Problem The description of the Super Grain case study in Section 3.1 ends with Clair Syverson (Super Grain’s vice president for marketing) sending a memorandum to the company’s president, David Sloan, requesting a meeting to evaluate her proposed promotional campaign for the company’s new breakfast cereal. Soon thereafter, Claire Syverson and David Sloan meet to discuss plans for the campaign. David Sloan (president): Thanks for your memo, Claire. The plan you outline for the promotional campaign looks like a reasonable one. However, I am surprised that it does not make any use of TV commercials. Why is that? Claire Syverson (vice president for marketing): Well, as I described in my memo, I used a spreadsheet model to see how to maximize the number of exposures from the campaign and this turned out to be the plan that does this. I also was surprised that it did not include TV commercials, but the model indicated that introducing commercials would provide less exposures on a dollar-for-dollar basis than magazine ads and Sunday supplement ads. Don’t you think it makes sense to use the plan that maximizes the number of exposures? David: Not necessarily. Some exposures are a lot less important than others. For example, we know that middle-aged adults are not big consumers of our cereals, so we don’t care very much how many of those people see our ads. On the other hand, young children are big consumers. Having TV commercials on the Saturday morning programs for children is our TABLE 3.6 Types of Functional Constraints Type Form* Typical Interpretation Main Usage Resource constraint LHS ≤ RHS Benefit constraint LHS ≥ RHS Resource-allocation problems and mixed problems Cost–benefit–trade-off problems and mixed problems Fixed-requirement constraint LHS = RHS For some resource, Amount used ≤ Amount available For some benefit, Level achieved ≥ Minimum acceptable level For some quantity, Amount provided = Required amount *LHS = Left-hand side (a SUMPRODUCT function). RHS = Right-hand side (a constant). Fixed-requirements problems and mixed problems 3.4 Mixed Problems 89 primary method of reaching young children. You know how important it will be to get young children to ask their parents for Crunchy Start. That is our best way of generating first-time sales. Those commercials also get seen by a lot of parents who are watching the programs with their kids. What we need is a commercial that is appealing to both parents and kids, and that gets the kids immediately bugging their parents to go buy Crunchy Start. I think that is a real key to a successful campaign. Claire: Yes, that makes a lot of sense. In fact, I already have set some goals regarding the number of young children and the number of parents of young children that need to be reached by this promotional campaign. David: Good. Did you include those goals in your spreadsheet model? Claire: No, I didn’t. David: Well, I suggest that you incorporate them directly into your model. I suspect that maximizing exposures while also meeting your goals will give us a high impact plan that includes some TV commercials. Claire: Good idea. I’ll try it. David: Are there any other factors that the plan in your memo doesn’t take into account as well as you would like? Claire: Well, yes, one. The plan doesn’t take into account my budget for cents-off coupons in magazines and newspapers. David: You should be able to add that to your model as well. Why don’t you go back and see what happens when you incorporate these additional considerations? Claire: OK, will do. You seem to have had a lot of experience with spreadsheet modeling. David: Yes. It is a great tool as long as you maintain some healthy skepticism about what comes out of the model. No model can fully take into account everything that we must consider when dealing with managerial problems. This is especially true the first time or two you run the model. You need to keep asking, what are the missing quantitative considerations that I still should add to the model? Then, after you have made the model as complete as possible and obtained a solution, you still need to use your best managerial judgment to weigh intangible considerations that cannot be incorporated into the model. Incorporating Additional Managerial Considerations into the Super Grain Model Therefore, David and Claire conclude that the spreadsheet model needs to be expanded to incorporate some additional considerations. In particular, since the promotional campaign is for a breakfast cereal that should have special appeal to young children, they feel that two audiences should be targeted—young children and parents of young children. (This is why one of the three advertising media recommended by Giacomi & Jackowitz is commercials on children’s television programs Saturday morning.) Consequently, Claire now has set two new goals for the campaign. Goal 1: The advertising should be seen by at least five million young children. Goal 2: The advertising should be seen by at least five million parents of young children. In effect, these two goals are minimum acceptable levels for two special benefits to be achieved by the advertising activities. Benefit 1: Promoting the new breakfast cereal to young children. Benefit 2: Promoting the new breakfast cereal to parents of young children. Because of the way the goals have been articulated, the level of each of these benefits is measured by the number of people in the specified category that are reached by the advertising. To enable constructing the corresponding benefit constraints (as described in Section 3.3), Claire asks Giacomi & Jackowitz to estimate how much each advertisement in each of the media will contribute to each benefit, as measured by the number of people reached in the specified category. These estimates are given in Table 3.7. 90 Chapter Three Linear Programming: Formulation and Applications TABLE 3.7 Benefit Data for the Revised Super Grain Corp. Advertising-Mix Problem Number Reached in Target Category (in millions) Each TV Commercial Each Magazine Ad Young children 1.2 0.1 0 5 Parents of young children 0.5 0.2 0.2 5 Target Category Benefit constraints are useful for incorporating managerial goals into the model. Each Sunday Ad Minimum Acceptable Level It is interesting to observe that management wants special consideration given to these two kinds of benefits even though the original spreadsheet model (Figure 3.1) already takes them into account to some extent. As described in Section 3.1, the expected number of exposures is the overall measure of performance to be maximized. This measure counts up all the times that an advertisement is seen by any individual, including all those individuals in the target audiences. However, maximizing this general measure of performance does not ensure that the two specific goals prescribed by management (Claire Syverson) will be achieved. Claire feels that achieving these goals is essential to a successful promotional campaign. Therefore, she complements the general objective with specific benefit constraints that do ensure that the goals will be achieved. Having benefit constraints added to incorporate managerial goals into the model is a prerogative of management. Claire has one more consideration she wants to incorporate into the model. She is a strong believer in the promotional value of cents-off coupons (coupons that shoppers can clip from printed advertisements to obtain a refund of a designated amount when purchasing the advertised item). Consequently, she always earmarks a major portion of her annual marketing budget for the redemption of these coupons. She still has $1,490,000 left from this year’s allotment for coupon redemptions. Because of the importance of Crunchy Start to the company, she has decided to use this entire remaining allotment in the campaign promoting this cereal. This fixed amount for coupon redemptions is a fixed requirement that needs to be expressed as a fixed-requirement constraint. As described at the beginning of this section, the form of a fixed-requirement constraint is that, for some type of quantity, Amount provided = Required amount In this case, the quantity involved is the amount of money provided for the redemption of cents-off coupons. To specify this constraint in the spreadsheet, we need to estimate how much each advertisement in each of the media will contribute toward fulfilling the required amount for the quantity. Both medium 2 (advertisements in food and family-oriented magazines) and medium 3 (advertisements in Sunday supplements of major newspapers) will feature cents-off coupons. The estimates of the amount of coupon redemption per advertisement in each of these media is given in Table 3.8. Formulation of the Revised Spreadsheet Model Figure 3.7 shows one way of formatting the spreadsheet to expand the original spreadsheet model in Figure 3.1 to incorporate the additional managerial considerations. We then outline the four components of the revised model next. TABLE 3.8 Data for the Fixed-Requirement Constraint for the Revised Super Grain Corp. Advertising-Mix Problem Contribution toward Required Amount Requirement Coupon redemption Each TV Spot Each Magazine Ad Each Sunday Ad Required Amount 0 $40,000 $120,000 $1,490,000 3.4 Mixed Problems 91 FIGURE 3.7 The spreadsheet model for the revised Super Grain problem, including the formulas for the objective cell TotalExposures (H19) and the other output cells in column F, as well as the specifications needed to set up Solver. The changing cells NumberOfAds (C19:E19) show the optimal solution obtained by Solver. A 1 B C D E F TV Spots Magazine Ads SS Ads 1,300 600 500 G H Super Grain Corp. Advertising-Mix Problem 2 3 4 Exposures per Ad 5 (thousands) Cost per Ad ($thousands) 6 Budget Spent Budget Available Ad Budget 300 150 100 3,775 ≤ 4,000 Planning Budget 90 30 40 1,000 ≤ 1,000 11 Young Children 1.2 0.1 0 5 ≥ 12 Parents of Young Children 0.5 0.2 0.2 5.85 ≥ TV Spots Magazine Ads SS Ads Total Redeemed 0 40 120 1,490 7 8 9 10 Number Reached per Ad (millions) Total Reached Minimum Acceptable 5 TotalExposures 5 ($thousands) 13 14 15 Coupon Redemption 16 per Ad ($thousands) Required Amount = 1,490 17 Total Exposures 18 19 Number of Ads Magazine Ads SS Ads (thousands) 3 14 7.75 16,175 ≤ 20 21 TV Spots Maximum TV Spots Solver Parameters Set Objective Cell: TotalExposures To: Max By Changing Variable Cells: NumberOfAds Subject to the Constraints: BudgetSpent <= Budget Available TVSpots <= MaxTVSpots TotalReached >= MinimumAcceptable TotalRedeemed = RequiredAmount Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 5 Range Name Cells BudgetAvailable BudgetSpent CostPerAd CouponRedemptionPerAd ExposuresPerAd MaxTVSpots MinimumAcceptable NumberOfAds NumberReachedPerAd RequiredAmount TotalExposures TotalReached TotalRedeemed TVSpots H7:H8 F7:F8 C7:E8 C15:E15 C4:E4 C21 H11:H12 C19:E19 C11:E12 H15 H19 F11:F12 F15 C19 F 6 Budget Spent 7 =SUMPRODUCT(C7:E7,NumberOfAds) 8 9 =SUMPRODUCT(C8:E8,NumberOfAds) 10 Total Reached 11 =SUMPRODUCT(C11:E11,NumberOfAds) 12 13 =SUMPRODUCT(C12:E12,NumberOfAds) 14 15 Total Redeemed =SUMPRODUCT(CouponRedemptionPerAd, NumberOfAds) H 17 18 19 Total Exposures (thousands) =SUMPRODUCT(ExposuresPerAd,NumberOfAds) The Data Additional data cells in NumberReachedPerAd (C11:E12), MinimumAcceptable (H11:H12), CouponRedemptionPerAd (C15:E15), and RequiredAmount (H15) give the data in Tables 3.7 and 3.8. 92 Chapter Three Linear Programming: Formulation and Applications The Decisions Recall that, as before, the decisions to be made are TV = Number of commercials on television M = Number of advertisements in magazines SS = Number of advertisements in Sunday supplements The changing cells to hold these numbers continue to be in NumberOfAds (C19:E19). The Constraints In addition to the original constraints, we now have two benefit constraints and one fixedrequirement constraint. As specified in rows 11 and 12, columns F to H, the benefit constraints are Total number of young children reached ≥ 5 (goal 1 in millions) Total number of parents reached ≥5 (goal 2 in millions) Using the data in columns C to E of these rows, Total number of young children reached = 1.2TV + 0.1M + 0SS = SUMPRODUCT (C11:E11, NumberOfAds) → cell F11 Total number of parents reached = 0.5TV + 0.2M + 0.2SS = SUMPRODUCT (C12:E12, NumberOfAds) → cell F12 These output cells are given the range name TotalReached (F11:F12). The fixed-requirement constraint indicated in row 15 is that Total coupon redemption = 1,490 (allotment in $1,000s) CouponRedemptionPerAd (C15:E15) gives the number of coupons redeemed per ad, so Total coupon redemption = 0TV + 40M + 120SS = SUMPRODUCT (CouponRedemptionPerAd, NumberOfAds) → cell F15 These same constraints are specified in Solver, along with the original constraints, in Figure 3.7. The Measure of Performance The measure of performance continues to be Exposure = 1,300TV + 600M + 500SS = SUMPRODUCT (ExposuresPerAd, NumberOfAds) → cell H19 so the objective cell is again TotalExposures (H19). Summary of the Formulation The above steps have resulted in formulating the following linear programming model (in algebraic form) on a spreadsheet. Maximize Exposure = 1,300TV + 600M + 500SS subject to the following constraints: 1. Resource constraints: 300TV + 150M + 100SS ≤ 4,000 (ad budget in $1,000s) ≤ 1,000 ) 90TV + 30M + 40SS ( planning budget in $1,000s TV ≤ 5 ( television spots available) An Application Vignette The Chevron Corporation is one of the world’s leading integrated energy companies. It explores extensively for crude oil and natural gas throughout the world. Thanks to its vast reserves, it produces nearly 2 million barrels of crude oil per day and similar amounts of natural gas. It then uses its refineries to refine and market nearly 3 million barrels per day of transportation fuels, chemicals, and lubricants. Dating nearly back to the invention of linear programming in 1947, Chevron quickly became one of the heaviest users of this exciting new technique. The earliest applications involved the following blending problem. Any single grade of gasoline needs to be blended from about three to ten components (different forms of processed crude oil), where no single component meets the quality specifications of the grade of gasoline but various combinations of the components can accomplish this. A typical refinery might have 20 different components to be blended into four or more grades of gasoline differing in octane and other properties by marketing area. Linear programming of the mixed type can achieve huge savings by solving for how to minimize the total cost of accomplishing all this blending. As time went on, the exponential growth in computing power enabled Chevron to greatly expand its use of linear programming. One application involves optimizing the combination of refined products (gasoline, jet, diesel fuels) to produce to maximize the total profit. Another application involves periodically determining the optimal way to run the refining processing units when changes occur in crude oil prices, raw material availability, product prices, product specifications, and equipment capabilities. Still another application of linear programming (along with also using decision analysis, the subject of Chapter 9) involves optimizing the use of capital for new projects to improve its refining system on an ongoing basis. The combination of all these applications of linear programming to minimize the total cost or maximize the total profit in various ways has had a dramatic impact on Chevron’s bottom line. The estimated cumulative value to Chevron now approaches $1 billion annually. In recognition of this and related work, the Institute for Operations Research and the Management Sciences (INFORMS) awarded Chevron the prestigious 2015 INFORMS Prize for its long and innovative history in applying advanced analytics and management science across the company. Source: T. Kutz, M. Davis, R. Creek, N. Kenaston, C. Stenstrom, and M. Connor, Interfaces 44, no. 1 (January-February 2014), pp. 39–54. (A link to this article is available at www.mhhe.com/Hillier6e.) 2. Benefit constraints: 1.2TV + 0.1M ≥5 (millions of young children) 0.5TV + 0.2M + 0.2SS ≥ 5 (millions of parents) 3. Fixed-requirement constraint: 40M + 120SS = 1,490 (coupon budget in $1,000s) 4. Nonnegativity constraints: V ≥ 0 T M ≥ 0 SS ≥ 0 Solving the Model The lower left-hand corner of Figure 3.7 shows the entries needed in Solver, along with the selection of the usual two options. Solver then finds the optimal solution given in row 19. This optimal solution provides the following plan for the promotional campaign: Run 3 television commercials. Run 14 advertisements in magazines. Run 7.75 advertisements in Sunday supplements (so the eighth advertisement would appear in only 75 percent of the newspapers). A model may need to be modified a number of times before it adequately incorporates all the important considerations. Although the expected number of exposures with this plan is only 16,175,000, versus the 17,000,000 with the first plan shown in Figure 3.1, both Claire Syverson and David Sloan feel that the new plan does a much better job of meeting all of management’s goals for this campaign. They decide to adopt the new plan. This case study illustrates a common theme in real applications of linear programming— the continuing evolution of the linear programming model. It is common to make later adjustments in the initial version of the model, perhaps even many times, as experience is gained in using the model. Frequently, these adjustments are made to more adequately reflect some important managerial considerations. This may result in a mixed problem because the new functional constraints needed to incorporate the managerial considerations may be of a different type from those in the original model. 93 94 Chapter Three Linear Programming: Formulation and Applications Other Examples This case study provides a relatively simple example of a small mixed problem. Most of the mixed problems that arise in practice are much larger, sometimes involving hundreds or thousands of activities and hundreds or thousands of constraints. At first glance, these larger problems may seem considerably more complicated than the case study. However, the important thing to remember is that any linear programming problem can have only three types of functional constraints—resource constraints, benefit constraints, and fixedrequirement ­constraints—where each type is formulated just as illustrated above for the case study. There are numerous kinds of managerial problems to which linear programming can be applied. We don’t have nearly enough space available to give examples of all the most important kinds of applications. However, if you would like to explore this further, we suggest that you go through the five solved problems that are summarized in front of the Problems section for this chapter. Reading the seven cases that follow the Problems section, as well as the application vignettes in both this chapter and the preceding chapter, also will further illustrate the unusually wide applicability of linear programming. Meanwhile, we soon will turn to two more categories of linear programming problems in the next two sections. Summary of the Formulation Procedure for Mixed Linear Programming Problems The procedure for formulating mixed problems is similar to the one outlined at the end of Section 3.2 for resource-allocation problems. However, pure resource-allocation prob­ lems only have resource constraints whereas mixed problems can include all three types of functional constraints (resource constraints, benefit constraints, and fixed-requirement constraints). Therefore, the following summary for formulating mixed problems includes separate steps for dealing with these different types of constraints. Also see Figure 3.8 for a template of the format for a spreadsheet model of mixed problems. (This format works well for most mixed problems, including those encountered in this chapter, but more flexibility is occasionally needed, as will be illustrated in the next chapter.) FIGURE 3.8 A template of a spreadsheet model for mixed problems. Activities Unit Profit or Cost Profit/cost per unit of activity Constraints Resource used per unit of activity SUMPRODUCT (resource used per unit, changing cells) Resources Available ≤ Benefit Achieved Benefit achieved per unit of activity Level of Activity Resources Used Changing cells SUMPRODUCT (benefit per unit, changing cells) Benefit Needed ≥ = Total Profit or Cost SUMPRODUCT(profit/cost per unit, changing cells) 3.5 Transportation Problems 95 1. Since any linear programming problem involves finding the best mix of levels of various activities, identify these activities for the problem at hand. The decisions to be made are the levels of these activities. 2. From the viewpoint of management, identify an appropriate overall measure of performance for solutions of the problem. 3. For each activity, estimate the contribution per unit of the activity to this overall measure of performance. 4. Identify any resources that must be allocated to the activities (as described in Section 3.2). For each one, identify the amount available and then the amount used per unit of each activity. 5. Identify any benefits to be obtained from the activities (as described in Section 3.3). For each one, identify the minimum acceptable level prescribed by management and then the benefit contribution per unit of each activity. 6. Identify any fixed requirements that, for some type of quantity, the amount provided must equal a required amount (as described in Section 3.4). For each fixed requirement, identify the required amount and then the contribution toward this required amount per unit of each activity. 7. Enter the data gathered in steps 3–6 into data cells in a spreadsheet. 8. Designate changing cells for displaying the decisions on activity levels. 9. Use output cells to specify the constraints on resources, benefits, and fixed requirements. 10. Designate an objective cell for displaying the overall measure of performance. Review Questions 3.5 1. What types of functional constraints can appear in a mixed linear programming problem? 2. What managerial goals needed to be incorporated into the expanded linear programming model for the Super Grain Corp. problem? 3. Which categories of functional constraints are included in the new linear programming model? 4. Why did management adopt the new plan even though it provides a smaller expected number of exposures than the original plan recommended by the original linear programming model? TRANSPORTATION PROBLEMS One of the most common applications of linear programming involves optimizing a shipping plan for transporting goods. In a typical application, a company has several plants producing a certain product that needs to be shipped to the company’s customers (or perhaps to distribution centers). How much should each plant ship to each customer in order to minimize the total cost? Linear programming can provide the answer. This type of linear programming problem is called a transportation problem. This kind of application normally needs two kinds of functional constraints. One kind specifies that the amount of the product produced at each plant must equal the total amount shipped to customers. The other kind specifies that the total amount received from the plants by each customer must equal the amount ordered. These are fixed-requirement constraints, which makes the problem a fixed-requirements problem. However, there also are variations of this problem where resource constraints or benefit constraints are needed. Transportation problems and assignment problems (described in the next section) are such important types of linear programming problems that the entire Chapter 15 (available at www.mhhe.com/Hillier6e) is devoted to further describing these two related types of problems and providing examples of a wide variety of applications. We provide below an example of a typical transportation problem. The Big M Company Transportation Problem The Big M Company produces a variety of heavy duty machines at two factories. One of its products is a large turret lathe. Orders have been received from three customers to purchase some of these turret lathes next month. These lathes will be shipped individually, and Table 3.9 shows what the cost will be for shipping each lathe from each factory to each customer. This 96 Chapter Three Linear Programming: Formulation and Applications TABLE 3.9 Some Data for the Big M Company DistributionNetwork Problem Shipping Cost for Each Lathe To Customer 1 Customer 2 Customer 3 Output From Factory 1 Factory 2 $700 800 $900 900 $800 700 12 lathes 15 lathes Order size 10 lathes 8 lathes 9 lathes table also shows how many lathes have been ordered by each customer and how many will be produced by each factory. The company’s distribution manager now wants to determine how many machines to ship from each factory to each customer to minimize the total shipping cost. Figure 3.9 depicts the distribution network for this problem. This network ignores the geographical layout of the factories and customers and instead lines up the two factories in one column on the left and the three customers in one column on the right. Each arrow shows one of the shipping lanes through this distribution network. Formulation of the Problem in Linear Programming Terms We need to identify the activities and requirements of this transportation problem to formulate it as a linear programming problem. In this case, two kinds of activities have been ­mentioned—the production of the turret lathes at the two factories and the shipping of these lathes along the various shipping lanes. However, we know the specific amounts to be produced at each factory, so no decisions need to be made about the production activities. The decisions to be made concern the levels of the shipping activities—how many lathes to ship through each shipping lane. Therefore, we need to focus on the shipping activities for the linear programming formulation. The activities correspond to shipping lanes, depicted by arrows in Figure 3.9. The level of each activity is the number of lathes shipped through the corresponding shipping lane. Just as any linear programming problem can be described as finding the best mix of activity levels, this one involves finding the best mix of shipping amounts for the various shipping lanes. The decisions to be made are SF1-C1 = Number of lathes shipped from Factory 1 to Customer 1 SF1-C2 = Number of lathes shipped from Factory 1 to Customer 2 FIGURE 3.9 The distribution network for the Big M Company problem. C1 10 lathes needed C2 8 lathes needed C3 9 lathes needed the 0/la $70 F1 $900/la the $8 00 /la the $8 00 /la th e 12 lathes produced 15 lathes produced F2 e /lath $900 $70 0 /lat he 3.5 Transportation Problems 97 SF1-C3 = Number of lathes shipped from Factory 1 to Customer 3 SF2-C1 = Number of lathes shipped from Factory 2 to Customer 1 SF2-C2 = Number of lathes shipped from Factory 2 to Customer 2 SF2-C3 = Number of lathes shipped from Factory 2 to Customer 3 so six changing cells will be needed in the spreadsheet. The objective is to Minimize Cost = Total cost for shipping the lathes Using the shipping costs given in Table 3.9, Cost = 700 S F1-C1 + 900 S F1-C2 + 800 S F1-C3 + 800 S F2-C1 + 900 S F2-C2 + 700 S F2-C3 is the quantity in dollars to be entered into the objective cell. (We will use a SUMPRODUCT function to do this a little later.) The spreadsheet model also will need five constraints involving fixed requirements. Both Table 3.9 and Figure 3.9 show these requirements. Requirement 1: Factory 1 must ship 12 lathes. Requirement 2: Factory 2 must ship 15 lathes. Requirement 3: Customer 1 must receive 10 lathes. Requirement 4: Customer 2 must receive 8 lathes. Requirement 5: Customer 3 must receive 9 lathes. Thus, there is a specific requirement associated with each of the five locations in the distribution network shown in Figure 3.9. All five of these requirements can be expressed in constraint form as Amount provided = Required amount For example, Requirement 1 can be expressed algebraically as S F1-C1 + S F1-C2 + S F1-C3 = 12 where the left-hand side gives the total number of lathes shipped from Factory 1, and 12 is the required amount to be shipped from Factory 1. Therefore, this constraint restricts SF1-C1, SF1-C2, and SF1-C3 to values that sum to the required amount of 12. In contrast to the ≤ form for resource constraints and the ≥ form for benefit constraints, the constraints express fixed requirements that must hold with equality, so this transportation problem falls into the category of fixed-requirements problems introduced in the preceding section. However, Chapter 15 (available at www.mhhe.com/Hillier6e) gives several examples that illustrate how variants of transportation problems can have resource constraints or benefit constraints as well. For example, if 12 lathes represent the manufacturing capacity of Factory 1 (the maximum number that can be shipped) rather than a requirement for how many must be shipped, the constraint just given for Requirement 1 would become a ≤ resource constraint instead. Such variations can be incorporated readily into the spreadsheet model. Careful problem formulation needs to precede model formulation. Here is an example where SUM functions are used for output cells instead of SUMPRODUCT functions. Formulation of the Spreadsheet Model In preparation for formulating the model, the problem has been formulated above by identifying the decisions to be made, the constraints on these decisions, and the overall measure of performance, as well as gathering all the important data displayed in Table 3.9. All this information leads to the spreadsheet model shown in Figure 3.10. The data cells include ShippingCost (C5:E6), Output (H11:H12), and OrderSize (C15:E15), incorporating all the data from Table 3.9. The changing cells are UnitsShipped (C11:E12), which give the decisions on the amounts to be shipped through the respective shipping lanes. The output cells are TotalShippedOut (F11:F12) and TotalToCustomer (C13:E13), where the SUM functions entered into these cells are shown below the spreadsheet in Figure 3.10. The constraints are that TotalShippedOut is required to equal Output and TotalToCustomer is required to equal OrderSize. These constraints have been specified on the spreadsheet and entered into Solver. The objective 98 Chapter Three Linear Programming: Formulation and Applications cell is TotalCost (H15), where its SUMPRODUCT function gives the total shipping cost. The lower left-hand corner of Figure 3.10 shows the entries needed in Solver, along with the selection of the usual two options. The layout of the spreadsheet is different than for all the prior linear programming examples in the book. Rather than a separate column for each activity and a separate row for each constraint, the cost data and changing cells are laid out in a table format. This format provides a more natural and compact way of displaying the constraints and results. UnitsShipped (C11:E12) in the spreadsheet in Figure 3.10 shows the result of applying Solver to obtain an optimal solution for the number of lathes to ship through each shipping lane. TotalCost (H15) indicates that the total shipping cost for this shipping plan is $20,500. Since any transportation problem is a special type of linear programming problem, it makes the standard assumption that fractional solutions are allowed. However, we actually don’t want this assumption for this particular application since only integer numbers of lathes can be shipped from a factory to a customer. Fortunately, even while making the standard assumption, the numbers in the optimal solution shown in UnitsShipped (C11:E12) only have integer values. This is no coincidence. Because of the form of its model, almost any transportation FIGURE 3.10 The spreadsheet model for the Big M Company problem, including the formulas for the objective cell TotalCost (H15) and the other output cells TotalShippedOut (F11:F12) and TotalToCustomer (C13:E13), as well as the specifications needed to set up Solver. The changing cells UnitsShipped (C11:E12) show the optimal solution obtained by Solver. A 1 B C D E F G H Big M Company Distribution Problem 2 3 Shipping Cost 4 (per Lathe) Customer 1 Customer 2 Customer 3 5 Factory 1 $700 $900 $800 6 Factory 2 $800 $900 $700 7 8 Total 9 Shipped Output 10 Units Shipped Customer 1 Customer 2 Customer 3 Out 11 Factory 1 12 Factory 2 10 0 2 6 0 9 12 15 13 Total to Customer 10 8 9 = = = Total Cost 10 8 9 $20,500 14 15 Order Size Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells: UnitsShipped Subject to the Constraints: TotalShippedOut = Output TotalToCustomer = OrderSize Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 13 Range Name Cells OrderSize Output ShippingCost TotalCost TotalShippedOut TotalToCustomer UnitsShipped C15:E15 H11:H12 C5:E6 H15 F11:F12 C13:E13 C11:E12 12 = = 15 F 8 Total 9 Shipped 10 11 =SUM(C11:E11) 12 =SUM(C12:E12) Out B C D E Total to Customer =SUM(C11:C12) =SUM(D11:D12) =SUM(E11:E12) H 14 15 Total Cost =SUMPRODUCT(ShippingCost,UnitsShipped) 3.6 Assignment Problems 99 problem (including this one) is guaranteed in advance to have an optimal solution that has only integer values despite the fact that fractional solutions also are allowed. In particular, as long as the data for the problem includes only integer values for all the supplies and demands (which are the outputs and order sizes in the Big M Company problem), any transportation problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its decision variables. Therefore, it is not necessary to add constraints to the model that require these variables to have only integer values. To summarize, here is the algebraic form of the linear programming model that has been formulated in the spreadsheet: Minimize Cost = 700 SF1-C1 + 900 SF1-C2 + 800 SF1-C3 + 800 SF2-C1 + 900 SF2-C2 + 700 SF2-C3 subject to the following constraints: 1. Fixed-requirement constraints: SF1-C1 + SF1-C2 + SF1-C3 = 12 (Factory 1) SF2-C1 + SF2-C2 + SF2-C3 = 15 (Factory 2) SF1-C1 + SF2-C1 SF1-C2 = 10 (Customer 1) + SF2-C2 SF1-C3 = 8 (Customer 2) + SF2-C3 = 9 (Customer 3) 2. Nonnegativity constraints: SF1-C1 ≥ 0 SF1-C2 ≥ 0 SF1-C3 ≥ 0 SF2-C1 ≥ 0 SF2-C2 ≥ 0 SF2-C3 ≥ 0 Review Questions 3.6 1. Why are transportation problems given this name? 2. What is an identifying feature of transportation problems? 3. How does the form of a fixed-requirement constraint differ from that of a resource constraint? A benefit constraint? 4. What are the quantities with fixed requirements in the Big M Company problem? ASSIGNMENT PROBLEMS We now turn to another special type of linear programming problem called assignment problems. As the name suggests, this kind of problem involves making assignments. Frequently, these are assignments of people to jobs. Thus, many applications of the assignment problem involve aiding managers in matching up their personnel with tasks to be performed. Other applications might instead involve assigning machines, vehicles, or plants to tasks. Here is a typical example. An Example: The Sellmore Company Problem The marketing manager of the Sellmore Company will be holding the company’s annual sales conference soon for sales regional managers and personnel. To assist in the administration of the conference, he is hiring four temporary employees (Ann, Ian, Joan, and Sean), where each will handle one of the following four tasks: 1. 2. 3. 4. Word processing of written presentations. Computer graphics for both oral and written presentations. Preparation of conference packets, including copying and organizing written materials. Handling of advance and on-site registrations for the conference. He now needs to decide which person to assign to each task. 100 Chapter Three Linear Programming: Formulation and Applications TABLE 3.10 Data for the Sellmore Co. Problem Required Time per Task (Hours) Temporary Employee Ann Ian Joan Sean Decisions need to be made regarding which person to assign to each task. Using Cost (D15:G18), the objective is to minimize the total cost of the assignments. A value of 1 in a changing cell indicates that the corresponding assignment is being made, whereas 0 means that the assignment is not being made. Excel Tip: When solving an assignment problem, rounding errors occasionally will cause Excel to return a noninteger value very close to 0 (e.g., 1.23 E-10, meaning 0.000000000123) or very close to 1 (e.g., 0.9999912). To make the spreadsheet cleaner, you may replace these “ugly” representations by their proper value of 0 or 1 in the changing cells. Word Processing 35 47 39 32 Graphics 41 45 56 51 Packets 27 32 36 25 Registrations 40 51 43 46 Hourly Wage $14 12 13 15 Although each temporary employee has at least the minimal background necessary to perform any of the four tasks, they differ considerably in how efficiently they can handle the different types of work. Table 3.10 shows how many hours each would need for each task. The rightmost column gives the hourly wage based on the background of each employee. Formulation of a Spreadsheet Model Figure 3.11 shows a spreadsheet model for this problem. Table 3.10 is entered at the top. Combining these required times and wages gives the cost (cells D15:G18) for each possible assignment of a temporary employee to a task, using equations shown at the bottom of Figure 3.11. This cost table is just the way that any assignment problem is displayed. The objective is to determine which assignments should be made to minimize the sum of the associated costs. The values of 1 in Supply (J24:J27) indicate that each person (assignee) listed in column C must perform exactly one task. The values of 1 in Demand (D30:G30) indicate that each task must be performed by exactly one person. These requirements then are specified in the constraints given in Solver. Each of the changing cells Assignment (D24:G27) is given a value of 1 when the corresponding assignment is being made, and a value of 0 otherwise. Therefore, the Excel equation for the objective cell, TotalCost = SUMPRODUCT(Cost, Assignment), gives the total cost for the assignments being made. The Solver Parameters box specifies that the goal is to minimize this objective cell. The changing cells in Figure 3.11 show the optimal solution obtained after running Solver. This solution is Assign Ann to prepare conference packets. Assign Ian to do the computer graphics. Assign Joan to handle registrations. Assign Sean to do the word processing. The total cost given in cell J30 is $1,957. Characteristics of Assignment Problems Note that all the functional constraints of the Sellmore Co. problem (as shown in cells H24:J27 and D28:G30 of Figure 3.11) are fixed-requirement constraints which require each person to perform exactly one task and require each task to be performed by exactly one person. Thus, like the Big M Company transportation problem, the Sellmore Co. is a fixed-requirements problem. This is a characteristic of all pure assignment problems. However, Chapter 15 (available at www.mhhe.com/Hillier6e) gives some examples of variants of assignment problems where this is not the case. Like the changing cells Assignment (D24:G27) in Figure 3.11, the changing cells in the spreadsheet model for any pure assignment problem gives a value of 1 when the corresponding assignment is being made and a value of 0 otherwise. Since the fixed-requirement constraints require only each row or column of changing cells to add up to 1 (which could happen, e.g., if two of the changing cells in the same row or column had a value of 0.5 and the rest 0), this would seem to necessitate adding the constraints that each of the changing cells must be integer. After choosing the Solver option to make the changing cells nonnegative, this then would force each of the changing cells to be 0 or 1. However, it turned out to be unnecessary to add the constraints that require the changing cells to have values of 0 or 1 3.6 Assignment Problems 101 FIGURE 3.11 A spreadsheet formulation of the Sellmore Co. problem as an assignment problem, including the objective cell TotalCost (J30) and the other output cells Cost (D15:G18), TotalAssignments (H24:H27), and TotalAssigned (D28:G28), as well as the specifications needed to set up the model. The values of 1 in the changing cells Assignment (D24:G27) show the optimal plan obtained by Solver for assigning the people to the tasks. A 1 2 B C D E F G H I J Sellmore Co. Assignment Problem Task 3 4 Required Time 5 (Hours) Word Hourly Processing Graphics Packets Registrations Wage Ann 35 41 27 40 $14 Ian 47 45 32 51 $12 8 Joan 39 56 36 43 $13 9 Sean 32 51 25 46 $15 6 Assignee 7 10 11 Task 12 Word 13 Processing Graphics Packets Registrations Ann $490 $574 $378 $560 Ian $564 $540 $384 $612 17 Joan $507 $728 $468 $559 18 Sean $480 $765 $375 $690 Cost 14 15 16 Assignee 19 20 Task 21 Word Assignment 22 Total Processing Graphics Packets Registrations Assignments Ann 0 0 1 0 1 = 1 Ian 0 1 0 0 1 = 1 26 Joan 0 0 0 1 1 = 1 27 Sean 1 0 0 0 1 = 1 28 Total Assigned 1 1 1 1 = = = 1 1 1 = 1 23 24 25 Assignee 29 Demand 30 B C D Cost 15 Ann F Total Cost $1,957 H G Word 22 Total Processing 23 Assignments =G6*I6 24 =SUM(D24:G24) =SUM(D25:G25) 13 14 E Supply =D6*I6 Graphics Packets Registrations =F6*I6 =E6*I6 Ian =D7*I7 =E7*I7 =F7*I7 =G7*I7 25 17 Joan =D8*I8 =E8*I8 =F8*I8 =G8*I8 26 =SUM(D26:G26) 18 Sean =D9*I9 =E9*I9 =F9*I9 =G9*I9 27 =SUM(D27:G27) 16 Assignee Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells: Assignment Subject to the Constraints: TotalAssigned = Demand TotalAssignments = Supply J 29 Total Cost 30 =SUMPRODUCT(Cost,Assignment) Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 28 C D E Total Assigned =SUM(D24:D27) =SUM(E24:E27) F G =SUM(F24:F27) =SUM(G24:G27) Range Name Cells Assignment Cost Demand HourlyWage RequiredTime Supply TotalAssigned TotalAssignments TotalCost D24:G27 D15:G18 D30:G30 I6:I9 D6:G9 J24:J27 D28:G28 H24:H27 J30 An Application Vignette Taylor Communications (formerly the Standard Operations Company before its acquisition and rebranding by the Taylor Corporation in 2016) is one of the largest printing conglomerates in the world. It provides high volume print production of forms and print stationery for major firms in and near the United States. It operates dozens of manufacturing facilities across the United States and Mexico. The print market is highly competitive, so company management must continually address the strategic challenge of minimizing the total cost of producing and distributing its products in order to offer competitive pricing. A wide variety of printing presses are available in the various manufacturing facilities and each press has the capacity to perform multiple print jobs. However, the cost of performing any given job can differ greatly from one press to the next. Furthermore, the cost of transporting the finished product to the customer can vary greatly depending on which manufacturing facility is used and the resulting distance to the customer. Given a large number of print jobs that need to be performed over the next time period, the problem is to assign the print jobs to presses. The cost of each individual assignment is the sum of the cost of producing that job on that press and the cost of transporting the finished job to the customer. The objective is to select all of the assignments of jobs to presses so as to minimize the total cost of all the assignments. This is in fact an assignment problem as described in the current section except for the modification that a given press might be able to handle multiple jobs. Therefore, a constraint is added for each press that the total processing time for all of the jobs being assigned to that press cannot exceed the total processing time that is available at that press. Periodically solving this very large assignment problem and implementing this solution has provided an estimated annual savings of over $10 million. Source: S. L. Ahire, M. F. Gorman, D. Dwiggins, and O. Mudry, “Operations Research Helps Reshape Operations Strategy at Standard Register Company,” Interfaces 37, no. 6 (November-December 2007), pp.553–565. (A link to this article is available at www.mhhe.com/Hillier6e.) in Figure 3.11 because Solver gave an optimal solution that had only values of 0 or 1 anyway. In fact, a general characteristic of pure assignment problems is that Solver always provides such an optimal solution without needing to add these additional constraints. As described further in Chapter 15, available at www.mhhe.com/Hillier6e, another interesting characteristic of any pure assignment problem is that it can be viewed as a special type of pure transportation problem. In particular, every fixed-requirement constraint in the corresponding transportation problem would require that either a row or column of changing cells add up to 1. This would result in Solver giving an optimal solution where every changing cell has a value of either 0 or 1, just as for the original assignment problem. Review Questions 3.7 1. Why are assignment problems given this name? 2. Pure assignment problems have what type of functional constraints? 3. What is the interpretation of the changing cells in the spreadsheet model of a pure assignment problem? MODEL FORMULATION FROM A BROADER PERSPECTIVE Formulating and analyzing a linear programming model provides information to help managers make their decisions. That means the model must accurately reflect the managerial view of the problem: Both the measure of performance and the constraints in a model need to reflect the managerial view of the problem. 102 • The overall measure of performance must capture what management wants accomplished. • When management limits the amounts of resources that will be made available to the activities under consideration, these limitations should be expressed as resource constraints. • When management establishes minimum acceptable levels for benefits to be gained from the activities, these managerial goals should be incorporated into the model as benefit constraints. • If management has fixed requirements for certain quantities, then fixed-requirement constraints are needed. With the help of spreadsheets, some managers now are able to formulate and solve small linear programming models themselves. However, larger linear programming models may be formulated by management science teams, not managers. When this is done, the management 3.7 Model Formulation from a Broader Perspective 103 Linear programming studies need strong managerial input and support. What-if analysis addresses some key questions that remain after formulating and solving a model. science team must thoroughly understand the managerial view of the problem. This requires clear communication with management from the very beginning of the study and maintaining effective communication as new issues requiring managerial guidance are identified. Management needs to clearly convey its view of the problem and the important issues involved. A manager cannot expect to obtain a helpful linear programming study without making clear just what help is wanted. As is necessary in any textbook, the examples in this chapter are far smaller, simpler, and more clearly spelled out than is typical of real applications. Many real studies require formulating complicated linear programming models involving hundreds or thousands (or possibly even millions) of decisions and constraints. In these cases, there usually are many ambiguities about just what should be incorporated into the model. Strong managerial input and support are vital to the success of a linear programming study for such complex problems. When dealing with huge real problems, there is no such thing as “the” correct linear programming model for the problem. The model continually evolves throughout the course of the study. Early in the study, various techniques are used to test initial versions of the model to identify the errors and omissions that inevitably occur when constructing such a large model. This testing process is referred to as model validation. Once the basic formulation has been validated, there are many reasonable variations of the model that might be used. Which variation to use depends on such factors as the assumptions about the problem that seem most reasonable, the estimates of the parameters of the model that seem most reliable, and the degree of detail desired in the model. In large linear programming studies, a good approach is to begin with a relatively simple version of the model and then use the experience gained with this model to evolve toward more elaborate models that more nearly reflect the complexity of the real problem. This process of model enrichment continues only as long as the model remains reasonably easy to solve. It must be curtailed when the study’s results are needed by management. Managers often need to curb the natural instinct of management science teams to continue adding “bells and whistles” to the model rather than winding up the study in a timely fashion with a less elegant but adequate model. When managers study the output of the current model, they often detect some undesirable characteristics that point toward needed model enrichments. These enrichments frequently take the form of new benefit constraints to satisfy some managerial goals not previously articulated. (Recall that this is what happened in the Super Grain case study.) Even though many reasonable variations of the model could be used, an optimal solution can be solved for only with respect to one specific version of the model at a time. This is why whatif analysis is such an important part of a linear programming study. After obtaining an optimal solution with respect to one specific model, management will have many what-if questions: • What if the estimates of the parameters in the model are incorrect? • How do the conclusions change if different plausible assumptions are made about the problem? • What happens when certain managerial options are pursued that are not incorporated into the current model? Chapter 5 is devoted primarily to describing how what-if analysis addresses these and related issues, as well as how managers use this information. Because managers instigate management science studies, they need to know enough about linear programming models and their formulation to be able to recognize managerial problems to which linear programming can be applied. Furthermore, since managerial input is so important for linear programming studies, managers need to understand the kinds of managerial concerns that can be incorporated into the model. Developing these two skills have been the most important goals of this chapter. Review Questions 1. 2. 3. 4. A linear programming model needs to reflect accurately whose view of the problem? What is meant by model validation? What is meant by the process of model enrichment? Why is what-if analysis an important part of a linear programming study? 104 Chapter Three Linear Programming: Formulation and Applications 3.8 Summary Functional constraints with a ≤ sign are called resource constraints, because they require that the amount used of some resource must be less than or equal to the amount available of that resource. The identifying feature of resource-allocation problems is that all their functional constraints are resource constraints. Functional constraints with a ≥ sign are called benefit constraints, since their form is that the level achieved for some benefit must be greater than or equal to the minimum acceptable level for that benefit. Frequently, benefit constraints express goals prescribed by management. If every functional constraint is a benefit constraint, then the problem is a cost–benefit–trade-off problem. Functional constraints with an = sign are called fixed-requirement constraints, because they express the fixed requirement that, for some quantity, the amount provided must be equal to the required amount. The identifying feature of fixed-requirements problems is that their functional constraints are fixed-requirement constraints. One prominent type of fixed-requirements problem is transportation problems, which typically involve finding a shipping plan that minimizes the total cost of transporting a product from a number of plants to a number of customers. Another prominent type is assignment problems, which typically involves assigning people to tasks so as to minimize the total cost of performing these tasks. Linear programming problems that do not fit into any of these three categories are called mixed problems. In many real applications, management science teams formulate and analyze large linear programming models to help guide managerial decision making. Such teams need strong managerial input and support to help ensure that their work really meets management’s needs. Glossary assignment problem A type of linear programming problem that typically involves assigning people to tasks so as to minimize the total cost of performing these tasks. (Section 3.6), 99 benefit constraint A functional constraint with a ≥ sign. The left-hand side is interpreted as the level of some benefit that is achieved by the activities under consideration, and the right-hand side is the minimum acceptable level for that benefit. (Section 3.3), 82 cost–benefit–trade-off problem A type of linear programming problem involving the trade-off between the total cost of the activities under consideration and the benefits to be achieved by these activities. Its identifying feature is that each functional constraint in the linear programming model is a benefit constraint. (Section 3.3), 82 fixed-requirement constraint A functional constraint with an = sign. The left-hand side represents the amount provided of some type of quantity, and the right-hand side represents the required amount for that quantity. (Section 3.4), 88 fixed-requirements problem A type of linear programming problem concerned with optimizing how to meet a number of fixed requirements. Its identifying feature is that each functional constraint in its model is a fixed-requirement constraint. (Section 3.4), 88 identifying feature A feature of a model that identifies the category of linear programming problem it represents. (Chapter introduction), 64 integer programming problem A variation of a linear programming problem that has the additional restriction that some or all of the decision variables must have integer values. (Section 3.2), 75 mixed problem Any linear programming problem that includes at least two of the three types of functional constraints (resource constraints, benefit constraints, and fixed-requirement constraints). (Section 3.4), 88 model enrichment The process of using experience with a model to identify and add important details that will provide a better representation of the real problem. (Section 3.7), 103 model validation The process of checking and testing a model to develop a valid model. (Section 3.7), 103 resource-allocation problem A type of linear programming problem concerned with allocating resources to activities. Its identifying feature is that each functional constraint in its model is a resource constraint. (Section 3.2), 71 resource constraint A functional constraint with a ≤ sign. The left-hand side represents the amount of some resource that is used by the activities under consideration, and the righthand side represents the amount available of that resource. (Section 3.2), 71 transportation problem A type of linear programming problem that typically involves finding a shipping plan that minimizes the total cost of transporting a product from a number of plants to a number of customers. (Section 3.5), 95 Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Excel Files: Super Grain Example TBA Airlines Example Think-Big Example Union Airways Example Big M Example Revised Super Grain Example Sellmore Example Excel Add-in: Analytic Solver Chapter 3 Solved Problems 105 Solved Problems The solutions are available at www.mhhe.com/Hillier6e. 3.S1. Farm Management Dwight and Hattie have run the family farm for over 30 years. They are currently planning the mix of crops to plant on their 120-acre farm for the upcoming season. The table gives the labor-hours and fertilizer required per acre, as well as the total expected profit per acre for each of the potential crops under consideration. Dwight, Hattie, and their children can work at most 6,500 total hours during the upcoming season. They have 200 tons of fertilizer available. What mix of crops should be planted to maximize the family’s total profit? a. Formulate and solve a linear programming model for this problem in a spreadsheet. b. Formulate this same model algebraically. Crop Labor Required (hours per acre) Fertilizer Required (tons per acre) Expected Profit (per acre) Oats Wheat Corn 50 60 105 1.5 2 4 $500 $600 $950 12-inch rods (with no waste). For the next production period, Decora needs twenty-five 12-inch rods, fifty-two 18-inch rods, forty-five 24-inch rods, thirty 40-inch rods, and twelve 60-inch rods. What is the fewest number of 60-inch rods that can be purchased to meet their production needs? Formulate and solve an integer programming model in a spreadsheet. 3.S4. Producing and Distributing AEDs at Heart Start Heart Start produces automated external defibrillators in each of two different plants (A and B). The unit production costs and monthly production capacity of the two plants are indicated in the table below. The automated external defibrillators are sold through three wholesalers. The shipping cost from each plant to the warehouse of each wholesaler along with the monthly demand from each wholesaler are also indicated in the table. The management of Heart Start now has asked their top management scientist (you) to address the following two questions. How many automated external defibrillators should be produced in each plant, and how should they be distributed to each of the three wholesaler warehouses so as to minimize the combined cost of production and shipping? Formulate and solve a linear programming model in a spreadsheet. Unit Shipping Cost 3.S2. Diet Problem The kitchen manager for Sing Sing prison is trying to decide what to feed its prisoners. She would like to offer some combination of milk, beans, and oranges. The goal is to minimize cost, subject to meeting the minimum nutritional requirements imposed by law. The cost and nutritional content of each food, along with the minimum nutritional requirements, are shown below. What diet should be fed to each prisoner? a. Formulate and solve a linear programming model for this problem in a spreadsheet. b. Formulate this same model algebraically. Oranges Navy (large Minimum Milk Beans Calif. Daily (gallons) (cups) Valencia) Requirement Niacin (mg) Thiamin (mg) Vitamin C (mg) Cost ($) 3.2 1.12 32.0 2.00 4.9 1.3 0.0 0.20 3.S3. Cutting Stock Problem 0.8 0.19 13.0 1.5 93.0 0.25 45.0 Decora Accessories manufactures a variety of bathroom accessories, including decorative towel rods and shower curtain rods. Each of the accessories includes a rod made out of stainless steel. However, many different lengths are needed: 12, 18, 24, 40, and 60 inches. Decora purchases 60-inch rods from an outside supplier and then cuts the rods as needed for their products. Each 60-inch rod can be used to make a number of smaller rods. For example, a 60-inch rod could be used to make a 40-inch and an 18-inch rod (with 2 inches of waste), or five Plant A Plant B Monthly Demand Unit Monthly Ware- Produc- Produchouse tion tion 3 Cost Capacity Warehouse 1 Warehouse 2 $22 $16 $14 $20 $30 $24 80 60 70 $600 $625 100 120 3.S5. Bidding for Classes In the MBA program at a prestigious university in the Pacific Northwest, students bid for electives in the second year of their program. Each student has 100 points to bid (total) and must take two electives. There are four electives available: Management Science (MS), Finance (Fin), Operations Management (OM), and Marketing (Mkt). Each class is limited to 5 students. The bids submitted for each of the 10 students are shown in the table below. Student Bids for Classes Student MS Fin OM Mkt George Fred Ann Eric Susan Liz Ed David Tony Jennifer 60 20 45 50 30 50 70 25 35 60 10 20 45 20 30 50 20 25 15 10 10 40 5 5 30 0 10 35 35 10 20 20 5 25 10 0 0 15 15 20 106 Chapter Three Linear Programming: Formulation and Applications a. Formulate and solve a spreadsheet model to determine an assignment of students to classes so as to maximize the total bid points of the assignments. b. Does the resulting solution seem like a fair assignment? c. Which alternative objectives might lead to a fairer assignment? Problems An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 3.1. Reconsider the Super Grain Corp. case study as presented in Section 3.1. The advertising firm, Giacomi & Jackowitz, now has suggested a fourth promising advertising medium—radio commercials—to promote the company’s new breakfast cereal, Crunchy Start. Young children are potentially major consumers of this cereal, but parents of young children (the major potential purchasers) often are too busy to do much reading (so may miss the company’s advertisements in magazines and Sunday supplements) or even to watch the Saturday morning programs for children where the company’s television commercials are aired. However, these parents do tend to listen to the radio during the commute to and from work. Therefore, to better reach these parents, Giacomi & Jackowitz suggests giving consideration to running commercials for Crunchy Start on nationally syndicated radio programs that appeal to young adults during typical commuting hours. Giacomi & Jackowitz estimates that the cost of developing each new radio commercial would be $50,000, and that the expected number of exposures per commercial would be 900,000. The firm has determined that 10 spots are available for different radio commercials, and each one would cost $200,000 for a normal run. a. Formulate and solve a spreadsheet model for the revised advertising-mix problem that includes this fourth advertising medium. Identify the data cells, the changing cells, and the objective cell. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. b. Indicate why this spreadsheet model is a linear programming model. c. Express this model in algebraic form. 3.2. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 3.2. Briefly describe how linear programming was applied in this study. Then list the various benefits that resulted from this study. 3.3.* Consider a resource-allocation problem having the following data. Resource Usage per Unit of Each Activity Resource 1 2 3 Contribution per unit 1 2 Amount of Resource Available 2 3 2 1 3 4 10 20 20 $20 $30 Contribution per unit = profit per unit of the activity. a. Formulate a linear programming model for this problem on a spreadsheet. b. Use the spreadsheet to check the following solutions: (x1, x2) = (2, 2), (3, 3), (2, 4), (4, 2), (3, 4), (4, 3). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? c. Use Solver to find an optimal solution. d. Express this model in algebraic form. e. Use the graphical method to solve this model. 3.4. Consider a resource-allocation problem having the following data. Resource Usage per Unit of Each Activity Resource 1 2 3 Amount of Resource Available A B C 30 0 20 20 10 20 0 40 30 500 600 1,000 $50 $40 $70 Contribution per unit Contribution per unit = profit per unit of the activity. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Express this model in algebraic form. 3.5. Consider a resource-allocation problem having the following data. Resource Usage per Unit of Each Activity Resource 1 2 3 4 Amount of Resource Available P Q R S 3 4 6 −2 5 −1 3 2 −2 3 2 5 4 2 −1 3 400 300 400 300 $11 $9 $8 $9 Contribution per unit Contribution per unit = profit per unit of the activity. a. Formulate a linear programming model for this problem on a spreadsheet. b. Make five guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value? c. Use Solver to find an optimal solution. Chapter 3 Problems 107 3.6.* The Omega Manufacturing Company has discontinued the production of a certain unprofitable product line. This act created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products, Products 1, 2, and 3. The available capacity of the machines that might limit output is summarized in the following table. Machine Type Available Time (in Machine-Hours per Week) Milling machine Lathe Grinder 500 350 150 The number of machine-hours required for each unit of the respective products are shown in the next table. Productivity Coefficient (in Machine-Hours per Unit) Machine Type Product 1 Product 2 Product 3 Milling machine Lathe Grinder 9 5 3 3 4 0 5 0 2 Each machine is available 40 hours per month. Each part manufactured will yield a unit profit as follows: Part Profit a. Indicate why this is a resource-allocation problem by identifying both the activities and the limited resources to be allocated to these activities. b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, the objective cell, and the other output cells. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve the model. e. Summarize the model in algebraic form. 3.7. Ed Butler is the production manager for the Bilco Corporation, which produces three types of spare parts for automobiles. The manufacture of each part requires processing on each of two machines, with the following processing times (in hours). Part Machine 1 2 A B C 0.02 0.05 0.03 0.02 0.05 0.04 B C $50 $40 $30 Ed wants to determine the mix of spare parts to produce to maximize total profit. a. Identify both the activities and the resources for this resource-allocation problem. b. Formulate a linear programming model for this problem on a spreadsheet. c. Make three guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value? d. Use Solver to find an optimal solution. e. Express the model in algebraic form. 3.8. Consider the following algebraic formulation of a resource-allocation problem with three resources, where the decisions to be made are the levels of three activities (A1, A2, and A3). Maximize The Sales Department indicates that the sales potential for Products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week. The unit profit would be $50, $20, and $25, respectively, for Products 1, 2, and 3. The objective is to determine how much of each product Omega should produce to maximize profit. A Profit = 20 A1 + 40 A2 + 30 A3 subject to Resource 1: 3 A1 + 5 A2 + 4 A3 ≤ 400 (amount available) Resource 2: A1 + A2 + A3 ≤ 100 (amount available) Resource 3: A1 + 3 A2 + 2 A3 ≤ 200 (amount available) and A1 ≥ 0 A2 ≥ 0 A3 ≥ 0 Formulate and solve the spreadsheet model for this problem. 3.9. Consider a cost–benefit–trade-off problem having the following data. Benefit Contribution per Unit of Each Activity Benefit 1 2 1 2 3 Unit cost 5 2 7 $60 3 2 9 $50 Minimum Acceptable Level 60 30 126 a. Formulate a linear programming model for this problem on a spreadsheet. b. Use the spreadsheet to check the following solutions: (x1, x2) = (7, 7), (7, 8), (8, 7), (8, 8), (8, 9), (9, 8). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? c. Use Solver to find an optimal solution. d. Express the model in algebraic form. e. Use the graphical method to solve this model. 108 Chapter Three Linear Programming: Formulation and Applications 3.10. Consider a cost–benefit–trade-off problem having the following data. Benefit Contribution per Unit of Each Activity Benefit 1 2 3 4 Minimum Acceptable Level P Q R 2 1 3 −1 4 5 4 −1 4 3 2 −1 80 60 110 Unit cost $400 $600 $500 $300 a. Formulate a linear programming model for this problem on a spreadsheet. b. Make five guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, for the value of the objective function. Which feasible guess has the best objective function value? c. Use Solver to find an optimal solution. 3.11.* Fred Jonasson manages a family-owned farm. To supplement several food products grown on the farm, Fred also raises pigs for market. He now wishes to determine the quantities of the available types of feed (corn, tankage, and alfalfa) that should be given to each pig. Since pigs will eat any mix of these feed types, the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with the daily nutritional requirements and feed costs. construction costs, Maureen needs to invest some of the company’s money now to meet these future cash flow needs. Maureen may purchase only three kinds of financial assets, each of which costs $1 million per unit. Fractional units may be purchased. The assets produce income 5, 10, and 20 years from now, and that income is needed to cover minimum cash flow requirements in those years, as shown in the following table. Income per Unit of Asset Year Asset 1 Asset 2 Asset 3 Minimum Cash Flow Required 5 10 20 $8 million 2 million 0 $4 million 2 million 6 million $2 million 4 million 8 million $1.6 billion 400 million 1.2 billion Maureen wishes to determine the mix of investments in these assets that will cover the cash flow requirements while minimizing the total amount invested. a. Formulate a linear programming model for this problem on a spreadsheet. b. Use the spreadsheet to check the possibility of purchasing 100 units of asset 1, 100 units of asset 2, and 200 units of asset 3. How much cash flow would this mix of investments generate 5, 10, and 20 years from now? What would be the total amount invested? c. Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the total amount invested for your solution? Kilogram of Corn Kilogram of Tankage Kilogram of Alfalfa Minimum Daily Requirement Carbohydrates Protein Vitamins 90 30 10 20 80 20 40 60 60 200 180 150 Cost (¢) 84 72 60 Nutritional Ingredient a. Formulate a linear programming model for this problem on a spreadsheet. b. Use the spreadsheet to check if (x1, x2, x3) = (1, 2, 2) is a feasible solution and, if so, what the daily cost would be for this diet. How many units of each nutritional ingredient would this diet provide daily? c. Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the daily cost for your solution? d. Use Solver to find an optimal solution. e. Express the model in algebraic form. 3.12. Maureen Laird is the chief financial officer for the Alva Electric Co., a major public utility in the Midwest. The company has scheduled the construction of new hydroelectric plants 5, 10, and 20 years from now to meet the needs of the growing population in the region served by the company. To cover the d. Use Solver to find an optimal solution. e. Summarize the model in algebraic form. 3.13. Web Mercantile sells many household products through an online catalog. The company needs substantial warehouse space for storing its goods. Plans now are being made for leasing warehouse storage space over the next five months. Just how much space will be required in each of these months is known. However, since these space requirements are quite different, it may be most economical to lease only the amount needed each month on a month-by-month basis. On the other hand, the additional cost for leasing space for additional months is much less than for the first month, so it may be less expensive to lease the maximum amount needed for the entire five months. Another option is the intermediate approach of changing the total amount of space leased (by adding a new lease and/or having an old lease expire) at least once but not every month. Chapter 3 Problems 109 The space requirement and the leasing costs for the various leasing periods are as follows. Month Required Space (Square Feet) 1 2 3 4 5 30,000 20,000 40,000 10,000 50,000 Leasing Period (Months) Cost per Sq. Ft. Leased 1 2 3 4 5 $ 65 100 135 160 190 The objective is to minimize the total leasing cost for meeting the space requirements. a. Indicate why this is a cost–benefit–trade-off problem by identifying both the activities and the benefits being sought from these activities. b. Identify verbally the decisions to be made, the constraints on these decisions, and the overall measure of performance for the decisions. c. Convert these verbal descriptions of the constraints and the measure of performance into quantitative expressions in terms of the data and decisions. d. Formulate a spreadsheet model for this problem. Identify the data cells, the changing cells, the objective cell, and the other output cells. Also show the Excel equation for each output cell expressed as a SUMPRODUCT function. Then use Solver to solve the model. e. Summarize the model in algebraic form. 3.14. Consider the following algebraic formulation of a cost– benefit–trade-off problem involving three benefits, where the decisions to be made are the levels of four activities (A1, A2, A3, and A4): Minimize Cost = 2 A1 + A2 − A3 + 3 A4 subject to Benefit 1: 3 A1 + 2 A2 − 2 A3 + 5 A4 ≥ 80 (minimum acceptable level) Benefit 2: A1 − A2 + A4 ≥ 10 (minimum acceptable level) Benefit 3: A1 + A2 − and A3 + 2 A4 ≥ 30 (minimum acceptable level) A1 ≥ 0 A2 ≥ 0 A3 ≥ 0 A4 ≥ 0 Formulate and solve the spreadsheet model for this problem. 3.15. Larry Edison is the director of the Computer Center for Buckly College. He now needs to schedule the staffing of the center. It is open from 8 AM until midnight. Larry has monitored the usage of the center at various times of the day and determined that the following number of computer consultants (qualified graduate students) are required. Minimum Number of Consultants Required to Be on Duty Time of Day 8 AM–noon Noon–4 PM 4 PM–8 PM 8 PM–midnight 6 8 12 6 Two types of computer consultants can be hired: full-time and part-time. The full-time consultants work for eight consecutive hours in any of the following shifts: morning (8 AM–4 PM), afternoon (noon–8 PM), and evening (4 PM–midnight). Full-time consultants are paid $17.50 per hour. Part-time consultants can be hired to work any of the four shifts listed in the table. Part-time consultants are paid $15 per hour. An additional requirement is that during every time period, there must be at least two full-time consultants on duty for every part-time consultant on duty. Larry would like to determine how many full-time and part-time consultants should work each shift to meet the above requirements at the minimum possible cost. a. Which category of linear programming problem does this problem fit? Why? b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Summarize the model in algebraic form. 3.16.* The Medequip Company produces precision medical diagnostic equipment at two factories. Three medical centers have placed orders for this month’s production output. The following table shows what the cost would be for shipping each unit from each factory to each of these customers. Also shown are the number of units that will be produced at each factory and the number of units ordered by each customer. A decision now needs to be made about the shipping plan for how many units to ship from each factory to each customer. a. Which category of linear programming problem does this problem fit? Why? b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Summarize this formulation in algebraic form. Unit Shipping Cost From To Customer Customer Customer 1 2 3 Output Factory 1 $600 $800 $700 Factory 2 400 900 600 300 units 200 units Order size 400 units 400 units 500 units 110 Chapter Three Linear Programming: Formulation and Applications 3.17. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 3.4. Briefly describe how linear programming was applied in this study. Then list the various benefits that resulted from this study. 3.18. The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron ore is shipped to either of two storage facilities. When needed, it then is shipped on to the company’s steel plant. The diagram below depicts this distribution network, where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel plant. The diagram also shows the monthly amounts produced at the mines and needed at the plant, as well as the shipping cost and the maximum amount that can be shipped per month through each shipping lane. 40 tons produced M1 $2,000/ton 30 tons max. S1 n /to x. 00 ma ,7 $1 ons t 30 70 M2 $1,100/ton 50 tons max. 100 tons needed P $ 50 1,60 to 0/t ns o m n ax . 60 tons produced $4 00 ton /ton sm ax . S2 n /to x. 00 $8 s ma ton 70 Management now wants to determine the most economical plan for shipping the iron ore from the mines through the distribution network to the steel plant. a. Identify all the requirements that will need to be expressed in fixed-requirement constraints. b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Express this model in algebraic form. 3.19.* Al Ferris has $60,000 that he wishes to invest now in order to use the accumulation for purchasing a retirement annuity in five years. After consulting with his financial advisor, he has been offered four types of fixed-income investments, which we will label as investments A, B, C, and D. Investments A and B are available at the beginning of each of the next five years (call them years 1 to 5). Each dollar invested in A at the beginning of a year returns $1.40 (a profit of $0.40) two years later (in time for immediate reinvestment). Each dollar invested in B at the beginning of a year returns $1.70 three years later. Investments C and D will each be available at one time in the future. Each dollar invested in C at the beginning of year 2 returns $1.90 at the end of year 5. Each dollar invested in D at the beginning of year 5 returns $1.30 at the end of year 5. Al wishes to know which investment plan maximizes the amount of money that can be accumulated by the beginning of year 6. a. For this problem, all its functional constraints can be expressed as fixed-requirement constraints. To do this, let At, Bt, Ct, and Dt be the amounts invested in investments A, B, C, and D, respectively, at the beginning of year t for each t where the investment is available and will mature by the end of year 5. Also let Rt be the number of available dollars not invested at the beginning of year t (and so available for investment in a later year). Thus, the amount invested at the beginning of year t plus Rt must equal the number of dollars available for investment at that time. Write such an equation in terms of the relevant variables above for the beginning of each of the five years to obtain the five fixed-requirement constraints for this problem. b. Formulate a complete linear programming model for this problem in algebraic form. c. Formulate and solve this model on a spreadsheet. 3.20. The Metalco Company desires to blend a new alloy of 40 percent tin, 35 percent zinc, and 25 percent lead from several available alloys having the following properties. Alloy Property 1 2 3 4 5 Percentage of tin Percentage of zinc Percentage of lead Cost ($/lb) 60 10 30 22 25 15 60 20 45 45 10 25 20 50 30 24 50 40 10 27 The objective is to determine the proportions of these alloys that should be blended to produce the new alloy at a minimum cost. a. Identify all the requirements that will need to be expressed in fixed-requirement constraints. b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Express this model in algebraic form. 3.21. The Weigelt Corporation has three branch plants with excess production capacity. Fortunately, the corporation has a new product ready to begin production, and all three plants have this capability, so some of the excess capacity can be used in this way. This product can be made in three sizes— large, medium, and small—that yield a net unit profit of $420, $360, and $300, respectively. Plants 1, 2, and 3 have the excess capacity to produce 750, 900, and 450 units per day of this product, respectively, regardless of the size or combination of sizes involved. The amount of available in-process storage space also imposes a limitation on the production rates of the new product. Plants 1, 2, and 3 have 13,000, 12,000, and 5,000 square feet, respectively, of in-process storage space available for a day’s production of this product. Each unit of the large, medium, and small sizes produced per day requires 20, 15, and 12 square feet, respectively. Sales forecasts indicate that if available, 900, 1,200, and 750 units of the large, medium, and small sizes, respectively, would be sold per day. At each plant, some employees will need to be laid off unless most of the plant’s excess production capacity can be used to produce the new product. To avoid layoffs if possible, management has decided that the plants should use the same percentage of their excess capacity to produce the new product. Management wishes to know how much of each of the sizes should be produced by each of the plants to maximize profit. a. Formulate and solve a linear programming model for this mixed problem on a spreadsheet. b. Express the model in algebraic form. Chapter 3 Problems 111 3.22.* A cargo plane has three compartments for storing cargo: front, center, and back. These compartments have capacity limits on both weight and space, as summarized below. Compartment Front Center Back Weight Capacity (Tons) Space Capacity (Cubic Feet) 12 18 10 7,000 9,000 5,000 Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that compartment’s weight capacity to maintain the balance of the airplane. The following four cargoes have been offered for shipment on an upcoming flight as space is available. Cargo Weight (Tons) Volume (Cubic Feet/Ton) Profit ($/Ton) 1 2 3 4 20 16 25 13 500 700 600 400 320 400 360 290 Any portion of these cargoes can be accepted. The objective is to determine how much (if any) of each cargo should be accepted and how to distribute each among the compartments to maximize the total profit for the flight. a. Formulate and solve a linear programming model for this mixed problem on a spreadsheet. b. Express the model in algebraic form. 3.23. Comfortable Hands is a company that features a product line of winter gloves for the entire family—men, women, and children. They are trying to decide what mix of these three types of gloves to produce. Comfortable Hands’s manufacturing labor force is unionized. Each full-time employee works a 40-hour week. In addition, by union contract, the number of full-time employees can never drop below 20. Nonunion, part-time workers also can be hired with the following union-imposed restrictions: (1) each part-time worker works 20 hours per week and (2) there must be at least two full-time employees for each parttime employee. All three types of gloves are made out of the same 100 percent genuine cowhide leather. Comfortable Hands has a long-term contract with a supplier of the leather and receives a 5,000-square-foot shipment of the material each week. The material requirements and labor requirements, along with the gross profit per glove sold (not considering labor costs), are given in the following table. Glove Men’s Women’s Children’s Material Required (Square Feet) Labor Required (Minutes) Gross Profit (per Pair) 2 1.5 1 30 45 40 $8 10 6 Each full-time employee earns $13 per hour, while each part-time employee earns $10 per hour. Management wishes to know what mix of each of the three types of gloves to produce per week, as well as how many full-time and part-time workers to employ. They would like to maximize their net profit—their gross profit from sales minus their labor costs. a. Formulate and solve a linear programming model for this problem on a spreadsheet. b. Summarize this formulation in algebraic form. 3.24. Oxbridge University maintains a powerful mainframe computer for research use by its faculty, Ph.D. students, and research associates. During all working hours, an operator must be available to operate and maintain the computer, as well as to perform some programming services. Beryl Ingram, the director of the computer facility, oversees the operation. It is now the beginning of the fall semester and Beryl is confronted with the problem of assigning different working hours to her operators. Because all the operators are currently enrolled in the university, they are available to work only a limited number of hours each day. There are six operators (four undergraduate students and two graduate students). They all have different wage rates because of differences in their experience with computers and in their programming ability. The following table shows their wage rates, along with the maximum number of hours that each can work each day. Maximum Hours of Availability Operators Wage Rate Mon. Tue. Wed. Thurs. Fri. K. C. $20.00/hour 6 0 6 0 6 D. H. $20.20/hour 0 6 0 6 0 H. B. $19.80/hour 4 8 4 0 4 S. C. $19.60/hour 5 5 5 0 5 K. S. $21.60/hour 3 0 3 8 0 N. K. $22.60/hour 0 0 0 6 2 Each operator is guaranteed a certain minimum number of hours per week that will maintain an adequate knowledge of the operation. This level is set arbitrarily at 8 hours per week for the undergraduate students (K. C., D. H., H. B., and S. C.) and 7 hours per week for the graduate students (K. S. and N. K.). The computer facility is to be open for operation from 8 AM to 10 PM Monday through Friday with exactly one operator on duty during these hours. On Saturdays and Sundays, the computer is to be operated by other staff. 112 Chapter Three Linear Programming: Formulation and Applications 12 grams (g) of protein. Furthermore, for practical reasons, each child needs exactly 2 slices of bread (to make the sandwich), at least twice as much peanut butter as jelly, and at least 1 cup of liquid (milk and/or juice). Joyce and Marvin would like to select the food choices for each child that minimize cost while meeting the above requirements. Because of a tight budget, Beryl has to minimize cost. She wishes to determine the number of hours she should assign to each operator on each day. Formulate and solve a spreadsheet model for this problem. 3.25. Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of its products is a strawberry shake that is designed to be a complete meal. The strawberry shake consists of several ingredients. Some information about each of these ingredients is given next. Ingredient Calories from Fat (per tbsp.) Total Calories (per tbsp.) Vitamin Content (mg/tbsp.) Thickeners (mg/tbsp.) Cost (¢/tbsp.) 1 75 0 0 30 50 100 0 120 80 20 0 50 0 2 3 8 1 2 25 10 8 25 15 6 Strawberry flavoring Cream Vitamin supplement Artificial sweetener Thickening agent The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No more than 20 percent of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin content. For taste reasons, there must be at least two tablespoons (tbsp.) of strawberry flavoring for each tbsp. of artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage. Management would like to select the quantity of each ingredient for the beverage that would minimize cost while meeting the above requirements. a. Identify the requirements that lead to resource constraints, to benefit constraints, and to fixed-requirement constraints. b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Summarize this formulation in algebraic form. 3.26. Joyce and Marvin run a day care for preschoolers. They are trying to decide what to feed the children for lunches. They would like to keep their costs down, but they also need to meet the nutritional requirements of the children. They have already decided to go with peanut butter and jelly sandwiches, and some combination of graham crackers, milk, and orange juice. The nutritional content of each food choice and its cost are given in the table below. Food Item Bread (1 slice) Peanut butter (1 tbsp.) Strawberry jelly (1 tbsp.) Graham cracker (1 cracker) Milk (1 cup) Juice (1 cup) a. Identify the requirements that lead to resource constraints, to benefit constraints, and to fixed-requirement constraints. b. Formulate and solve a linear programming model for this problem on a spreadsheet. c. Express the model in algebraic form. 3.27. The Cost-Less Corp. supplies its four retail outlets from its four plants. The shipping cost per shipment from each plant to each retail outlet is given below. Unit Shipping Cost Retail Outlet: Plant 1 2 3 4 1 2 3 4 $500 200 300 200 $600 900 400 100 $400 100 200 300 $200 300 100 200 Plants 1, 2, 3, and 4 make 10, 20, 20, and 10 shipments per month, respectively. Retail outlets 1, 2, 3, and 4 need to receive 20, 10, 10, and 20 shipments per month, respectively. Calories from Fat Total Calories Vitamin C (mg) Protein (g) Cost (¢) 10 75 0 20 70 0 70 100 50 60 150 100 0 0 3 0 2 120 3 4 0 1 8 1 5 4 7 8 15 35 The nutritional requirements are as follows. Each child should receive between 400 and 600 calories. No more than 30 percent of the total calories should come from fat. Each child should consume at least 60 milligrams (mg) of vitamin C and The distribution manager, Randy Smith, now wants to determine the best plan for how many shipments to send from each plant to the respective retail outlets each month. Randy’s objective is to minimize the total shipping cost. Chapter 3 Problems 113 Formulate this problem as a transportation problem on a spreadsheet and then use Solver to obtain an optimal solution. 3.28. The Childfair Company has three plants producing child push chairs that are to be shipped to four distribution centers. Plants 1, 2, and 3 produce 12, 17, and 11 shipments per month, respectively. Each distribution center needs to receive 10 shipments per month. The distance from each plant to the respective distribution centers is given below. Distance to Distribution Center (Miles) Plant 1 2 3 1 2 3 4 800 1,100 600 1,300 1,400 1,200 400 600 800 700 1,000 900 The freight cost for each shipment is $100 plus 50 cents/mile. How much should be shipped from each plant to each of the distribution centers to minimize the total shipping cost? Formulate this problem as a transportation problem on a spreadsheet and then use Solver to obtain an optimal solution. 3.29. The Onenote Co. produces a single product at three plants for four customers. The three plants will produce 60, 80, and 40 units, respectively, during the next week. The firm has made a commitment to sell 40 units to customer 1, 60 units to customer 2, and at least 20 units to customer 3. Both customers 3 and 4 also want to buy as many of the remaining units as possible. The net profit associated with shipping a unit from plant i for sale to customer j is given by the following table. Customer Plant 1 2 3 1 2 3 4 $800 500 600 $700 200 400 $500 100 300 $200 300 500 Management wishes to know how many units to sell to customers 3 and 4 and how many units to ship from each of the plants to each of the customers to maximize profit. Formulate and solve a spreadsheet model for this problem. 3.30. The Move-It Company has two plants building forklift trucks that then are shipped to three distribution centers. The production costs are the same at the two plants, and the cost of shipping each truck is shown below for each combination of plant and distribution center. Distribution Center Plant A B 1 2 3 $800 600 $700 800 $400 500 A total of 60 forklift trucks are produced and shipped per week. Each plant can produce and ship any amount up to a maximum of 50 trucks per week, so there is considerable flexibility on how to divide the total production between the two plants so as to reduce shipping costs. However, each distribution center must receive exactly 20 trucks per week. Management’s objective is to determine how many forklift trucks should be produced at each plant, and then what the overall shipping pattern should be to minimize total shipping cost. Formulate and solve a spreadsheet model for this problem. 3.31. Redo Problem 3.30 when any distribution center may receive any quantity between 10 and 30 forklift trucks per week in order to further reduce total shipping cost, provided only that the total shipped to all three distribution centers must still equal 60 trucks per week. 3.32. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 3.6. Briefly describe how the model for the assignment problem was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 3.33. Consider the assignment problem having the following cost table. Job Person A B C 1 2 3 $5 3 2 $7 6 3 $4 5 4 The optimal solution is A-3, B-1, C-2, with a total cost of $10. Formulate this problem on a spreadsheet and then use Solver to obtain the optimal solution identified above. 3.34. Four cargo ships will be used for shipping goods from one port to four other ports (labeled 1, 2, 3, 4). Any ship can be used for making any one of these four trips. However, because of differences in the ships and cargoes, the total cost of loading, transporting, and unloading the goods for the different ship–port combinations varies considerably, as shown in the following table. Port Ship 1 2 3 4 1 2 3 4 $500 600 700 500 $400 600 500 400 $600 700 700 600 $700 500 600 600 The objective is to assign the four ships to four different ports in such a way as to minimize the total cost for all four shipments. a. Describe how this problem fits into the format for an assignment problem. b. Formulate and solve this problem on a spreadsheet. 3.35. Reconsider Problem 3.30. Now distribution centers 1, 2, and 3 must receive exactly 10, 20, and 30 units per week, 114 Chapter Three Linear Programming: Formulation and Applications respectively. For administrative convenience, management has decided that each distribution center will be supplied totally by a single plant, so that one plant will supply one distribution center and the other plant will supply the other two distribution centers. The choice of these assignments of plants to distribution centers is to be made solely on the basis of minimizing total shipping cost. Formulate and solve a spreadsheet model for this problem. 3.36. Vincent Cardoza is the owner and manager of a machine shop that does custom order work. This Wednesday afternoon, he has received calls from two customers who would like to place rush orders. One is a trailer hitch company that would like some custom-made heavy-duty tow bars. The other is a minicar-carrier company that needs some customized stabilizer bars. Both customers would like as many as possible by the end of the week (two working days). Since both products would require the use of the same two machines, Vincent needs to decide and inform the customers this afternoon about how many of each product he will agree to make over the next two days. Each tow bar requires 3.2 hours on machine 1 and 2 hours on machine 2. Each stabilizer bar requires 2.4 hours on machine 1 and 3 hours on machine 2. Machine 1 will be available for 16 hours over the next two days and machine 2 will be available for 15 hours. The profit for each tow bar produced would be $130 and the profit for each stabilizer bar produced would be $150. Vincent now wants to determine the mix of these production quantities that will maximize the total profit. a. Formulate an integer programming model in algebraic form for this problem. b. Formulate and solve the model on a spreadsheet. 3.37. Pawtucket University is planning to buy new copier machines for its library. Three members of its Management Science Department are analyzing what to buy. They are considering two different models: Model A, a high-speed copier, and Model B, a lower speed but less expensive copier. Model A can handle 20,000 copies a day and costs $6,000. Model B can handle 10,000 copies a day but only costs $4,000. They would like to have at least six copiers so that they can spread them throughout the library. They also would like to have at least one high-speed copier. Finally, the copiers need to be able to handle a capacity of at least 75,000 copies per day. The objective is to determine the mix of these two copiers that will handle all these requirements at minimum cost. a. Formulate and solve a spreadsheet model for this problem. b. Formulate this same model in algebraic form. 3.38. Northeastern Airlines is considering the purchase of new long-, medium-, and short-range jet passenger airplanes. The purchase price would be $201 million for each longrange plane, $150 million for each medium-range plane, and $105 million for each short-range plane. The board of directors has authorized a maximum commitment of $4.5 billion for these purchases. Regardless of which airplanes are purchased, air travel of all distances is expected to be sufficiently large that these planes would be utilized at essentially maximum capacity. It is estimated that the net annual profit (after capital recovery costs are subtracted) would be $12.6 million per long-range plane, $9 million per medium-range plane, and $6.9 million per short-range plane. It is predicted that enough trained pilots will be available to the company to crew 30 new airplanes. If only short-range planes were purchased, the maintenance facilities would be able to handle 40 new planes. However, each medium-range plane is 11 equivalent to ___ short-range planes, and each long-range plane 3 12 is equivalent to ___ short-range planes in terms of their use of the 3 maintenance facilities. The information given here was obtained by a preliminary analysis of the problem. A more detailed analysis will be conducted subsequently. However, using the preceding data as a first approximation, management wishes to know how many planes of each type should be purchased to maximize profit. a. Formulate and solve a spreadsheet model for this problem. b. Formulate this model in algebraic form. Case 3-1 Shipping Wood to Market Alabama Atlantic is a lumber company that has three sources of wood and five markets to be supplied. The annual availability of wood at sources 1, 2, and 3 is 15, 20, and 15 million board feet, respectively. The amount that can be sold annually at markets 1, 2, 3, 4, and 5 is 11, 12, 9, 10, and 8 million board feet, respectively. Unit Cost by Rail ($1,000s) to Market Source 1 2 3 In the past, the company has shipped the wood by train. However, because shipping costs have been increasing, the alternative of using ships to make some of the deliveries is being investigated. This alternative would require the company to invest in some ships. Except for these investment costs, the Unit Cost by Ship ($1,000s) to Market 1 2 3 4 5 1 2 3 4 5 61 69 59 72 78 66 45 60 63 55 49 61 66 56 47 31 36 — 38 43 33 24 28 36 — 24 32 35 31 26 Case 3-2 Capacity Concerns 115 shipping costs in thousands of dollars per million board feet by rail and by water (when feasible) is given by the above table for each route. The capital investment (in thousands of dollars) in ships required for each million board feet to be transported annually by ship along each route is given next. Unit Investment for Ships ($1,000s) to Market Source 1 2 3 4 5 1 2 3 275 293 — 303 318 283 238 270 275 — 250 268 285 265 240 Considering the expected useful life of the ships and the time value of money, the equivalent uniform annual cost of these investments is one-tenth the amount given in the table. The objective is to determine the overall shipping plan that minimizes the total equivalent uniform annual cost (including shipping costs). You are the head of the management science team that has been assigned the task of determining this shipping plan for each of the three options listed next. Option 1: Continue shipping exclusively by rail. Option 2: Switch to shipping exclusively by water (except where only rail is feasible). Option 3: Ship by either rail or water, depending on which is less expensive for the particular route. Present your results for each option. Compare. Finally, consider the fact that these results are based on current shipping and investment costs, so that the decision on the option to adopt now should take into account management’s projection of how these costs are likely to change in the future. For each option, describe a scenario of future cost changes that would justify adopting that option now. Case 3-2 Capacity Concerns Bentley Hamilton throws the business section of The New York Times onto the conference room table and watches as his associates jolt upright in their overstuffed chairs. Mr. Hamilton wants to make a point. He throws the front page of the The Wall Street Journal on top of The New York Times and watches as his associates widen their eyes once heavy with boredom. Mr. Hamilton wants to make a big point. He then throws the front page of the Financial Times on top of the newspaper pile and watches as his associates dab the fine beads of sweat off their brows. Mr. Hamilton wants his point indelibly etched into his associates’ minds. “I have just presented you with three leading financial newspapers carrying today’s top business story,” Mr. H ­ amilton declares in a tight, angry voice. “My dear associates, our company is going to hell in a hand basket! Shall I read you the headlines? From The New York Times, ‘CommuniCorp stock drops to lowest in 52 weeks.’ From The Wall Street Journal, ‘CommuniCorp loses 25 percent of the wireless router market in only one year.’ Oh, and my favorite, from the Financial Times, ‘CommuniCorp cannot CommuniCate: CommuniCorp stock drops because of internal communications disarray.’ How did our company fall into such dire straits?” Mr. Hamilton next points at a line sloping slightly upward on the conference room display. “This is a graph of our productivity over the last 12 months. As you can see from the graph, productivity in our router production facility has increased steadily over the last year. Clearly, productivity is not the cause of our problem.” Mr. Hamilton next displays a second graph showing a line sloping steeply upward. “This is a graph of our missed or late orders over the last 12 months.” Mr. Hamilton hears an audible gasp from his associates. “As you can see from the graph, our missed or late orders have increased steadily and significantly over the past 12 months. I think this trend explains why we have been losing market share, causing our stock to drop to its lowest level in 52 weeks. We have angered and lost the business of retailers, our customers who depend upon on-time deliveries to meet the demand of consumers.” “Why have we missed our delivery dates when our productivity level should have allowed us to fill all orders?” Mr. ­Hamilton asks. “I called several departments to ask this question.” “It turns out that we have been producing routers for the hell of it!” Mr. Hamilton says in disbelief. “The marketing and sales departments do not communicate with the manufacturing department, so manufacturing executives do not know what routers to produce to fill orders. The manufacturing executives want to keep the plant running, so they produce routers regardless of whether the routers have been ordered. Finished routers are sent to the warehouse, but marketing and sales executives do not know the number and styles of routers in the warehouse. They try to communicate with warehouse executives to determine if the routers in inventory can fill the orders, but they rarely receive answers to their questions.” Mr. Hamilton pauses and looks directly at his associates. “Ladies and gentlemen, it seems to me that we have a serious internal communications problem. I intend to correct this problem immediately. I want to begin by installing a companywide computer network to ensure that all departments have access to critical documents and are able to communicate with each other more easily. Because this intranet will represent a large change from the current communications infrastructure, I expect some bugs in the system and some resistance from employees. I therefore want to phase in the installation of the intranet.” Mr. Hamilton passes the following time line and requirements chart to his associates (IN = intranet). 116 Chapter Three Linear Programming: Formulation and Applications Month 1 Month 2 Month 3 Month 4 Month 5 IN education Install IN in sales Install IN in manufacturing Install IN in warehouse Install IN in marketing Department Number of Employees Sales Manufacturing Warehouse Marketing 60 200 30 75 Mr. Hamilton proceeds to explain the time line and requirements chart. “In the first month, I do not want to bring any department onto the intranet; I simply want to disseminate information about it and get buy-in from employees. In the second month, I want to bring the sales department onto the intranet since the sales department receives all critical information from customers. In the third month, I want to bring the manufacturing department onto the intranet. In the fourth month, I want to install the intranet at the warehouse, and in the fifth and final month, I want to bring the marketing department onto the intranet. The requirements chart above lists the number of employees requiring access to the intranet in each department.” Mr. Hamilton turns to Emily Jones, the head of Corporate Information Management. “I need your help in planning for the installation of the intranet. Specifically, the company needs to purchase servers for the internal network. Employees will connect to company servers and download information to their own desktop computers.” Mr. Hamilton passes Emily the following chart detailing the types of servers available, the number of employees each server supports, and the cost of each server. Type of Server Number of Employees Server Supports Mini Desktop Server Desktop Server Workstation Server Full Rack Server Up to 30 employees Up to 80 employees Up to 200 employees Up to 2,000 employees “Emily, I need you to decide what servers to purchase and when to purchase them to minimize cost and to ensure that the company possesses enough server capacity to follow the intranet implementation timeline,” Mr. Hamilton says. “For example, you may decide to buy one large server during the first month to support all employees, or buy several small servers during the first month to support all employees, or buy one small server each month to support each new group of employees gaining access to the intranet.” “There are several factors that complicate your decision,” Mr. Hamilton continues. We can get a discount of 10 percent off each workstation server purchased, but only if we purchase servers in the first or second month. We can get a 25 percent discount off all full rack servers purchased in the first two months. You are also limited in the amount of money you can spend during the first month. CommuniCorp has already allocated much of the budget for the next two months, so you only have a total of $9,500 available to purchase servers in months 1 and 2. Finally, the manufacturing department requires at least one of the three more powerful servers. Have your decision on my desk at the end of the week.” a. Emily first decides to evaluate the number and type of servers to purchase on a month-to-month basis. For each month, formulate a spreadsheet model to determine which servers Emily should purchase in that month to minimize costs in that month and support the new users given your results for the preceding months. How many and which types of servers should she purchase in each month? How much is the total cost of the plan? b. Emily realizes that she could perhaps achieve savings if she bought a larger server in the initial months to support users in the final months. She therefore decides to evaluate the number and type of servers to purchase over the entire planning period. Formulate a spreadsheet model to determine which servers Emily should purchase in which months to minimize total cost and support all new users. How many and which types of servers should she purchase in each month? How much is the total cost of the plan? Cost of Server $ 2,500 5,000 10,000 25,000 c. Why is the answer using the first method different from that using the second method? d. Are there other costs for which Emily is not accounting in her problem formulation? If so, what are they? e. What further concerns might the various departments of CommuniCorp have regarding the intranet? Case 3-3 Fabrics and Fall Fashions 117 Case 3-3 Fabrics and Fall Fashions From the 10th floor of her office building, Katherine Rally watches the swarms of New Yorkers fight their way through the streets infested with yellow cabs and the sidewalks littered with hot dog stands. On this sweltering July day, she pays particular attention to the fashions worn by the various women and wonders what they will choose to wear in the fall. Her thoughts are not simply random musings; they are critical to her work since she owns and manages TrendLines, an elite women’s clothing company. Today is an especially important day because she must meet with Ted Lawson, the production manager, to decide upon next month’s production plan for the fall line. Specifically, she must determine the quantity of each clothing item she should produce given the plant’s production capacity, limited resources, and demand forecasts. Accurate planning for next month’s production is critical to fall sales since the items produced next month will appear in stores during September and women generally buy the majority of the fall fashions when they first appear in September. She turns back to her sprawling glass desk and looks at the numerous papers covering it. Her eyes roam across the clothing patterns designed almost six months ago, the lists of material requirements for each pattern, and the lists of demand forecasts for each pattern determined by customer surveys at fashion shows. She remembers the hectic and sometimes nightmarish days of designing the fall line and presenting it at fashion shows in New York, Milan, and Paris. Ultimately, she paid her team of six designers a total of $860,000 for their work on her fall line. With the cost of hiring runway models, hair stylists, and make-up artists; sewing and fitting clothes; building the set; choreographing and rehearsing the show; and renting the conference hall, each of the three fashion shows cost her an additional $2,700,000. She studies the clothing patterns and material requirements. Her fall line consists of both professional and casual fashions. She determined the price for each clothing item by taking into account the quality and cost of material, the cost of labor and machining, the demand for the item, and the prestige of the TrendLines brand name. The fall professional fashions include. Clothing Item Material Requirements Price Labor and Machine Cost Tailored wool slacks 3 yards of wool 2 yards of acetate for lining $300 $160 Cashmere sweater 1.5 yards of cashmere 450 150 Silk blouse Silk camisole 1.5 yards of silk 180 100 0.5 yard of silk 120 Tailored skirt 60 2 yards of rayon 270 120 320 140 1.5 yards of acetate for lining Wool blazer 2.5 yards of wool 1.5 yards of acetate for lining The fall casual fashions include. Labor and Machine Cost Clothing Item Material Requirements Price Velvet pants 3 yards of velvet $350 $175 2 yards of acetate for lining Cotton sweater 1.5 yards of cotton 130 60 Cotton miniskirt 0.5 yard of cotton 75 40 Velvet shirt 1.5 yards of velvet 200 160 Button-down blouse 1.5 yards of rayon 120 90 She knows that for the next month, she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet, and 30,000 yards of cotton for production. The prices of the materials are listed below. Material Wool Acetate Cashmere Silk Rayon Velvet Cotton Price per Yard $ 9.00 1.50 60.00 13.00 2.25 12.00 2.50 Any material that is not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler. She knows that the production of both the silk blouse and cotton sweater leaves leftover scraps of material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively, are needed. From these 2 yards, 1.5 yards are used for the silk blouse or the cotton sweater and 0.5 yard is left as scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton miniskirt, respectively. Therefore, whenever a silk blouse is produced, a silk camisole is also produced. Likewise, whenever a cotton sweater is produced, a cotton miniskirt is also produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton miniskirt without producing a cotton sweater. The demand forecasts indicate that some items have limited demand. Specifically, because the velvet pants and velvet shirts are fashion fads, TrendLines has forecasted that it can sell only 5,500 pairs of velvet pants and 6,000 velvet shirts. TrendLines does not want to produce more than the forecasted demand because once the pants and shirts go out of style, the 118 Chapter Three Linear Programming: Formulation and Applications company cannot sell them. TrendLines can produce less than the forecasted demand, however, since the company is not required to meet the demand. The cashmere sweater also has limited demand because it is quite expensive, and TrendLines knows it can sell at most 4,000 cashmere sweaters. The silk blouses and camisoles have limited demand because many women think silk is too hard to care for, and TrendLines projects that it can sell at most 12,000 silk blouses and 15,000 silk camisoles. The demand forecasts also indicate that the wool slacks, tailored skirts, and wool blazers have a great demand because they are basic items needed in every professional wardrobe. Specifically, the demand is 7,000 pairs of wool slacks and 5,000 wool blazers. Katherine wants to meet at least 60 percent of the demand for these two items to maintain her loyal customer base and not lose business in the future. Although the demand for tailored skirts could not be estimated, Katherine feels she should make at least 2,800 of them. a. Ted is trying to convince Katherine not to produce any velvet shirts since the demand for this fashion fad is quite low. He argues that this fashion fad alone accounts for $500,000 of the fixed design and other costs. The net contribution (price of clothing item – materials cost – labor cost) from selling the fashion fad should cover these fixed costs. Each velvet shirt generates a net contribution of $22. He argues that given the net contribution, even satisfying the maximum demand will not yield a profit. What do you think of Ted’s argument? b. Formulate and solve a linear programming problem to maximize profit given the production, resource, and demand constraints. Before she makes her final decision, Katherine plans to explore the following questions independently, except where otherwise indicated. c. The textile wholesaler informs Katherine that the velvet cannot be sent back because the demand forecasts show that the demand for velvet will decrease in the future. Katherine can therefore get no refund for the velvet. How does this fact change the production plan? d. What is an intuitive economic explanation for the difference between the solutions found in parts b and c? e. The sewing staff encounters difficulties sewing the arms and lining into the wool blazer since the blazer pattern has an awkward shape and the heavy wool material is difficult to cut and sew. The increased labor time to sew a wool blazer increases the labor and machine cost for each blazer by $80. Given this new cost, how many of each clothing item should TrendLines produce to maximize profit? f. The textile wholesaler informs Katherine that since another textile customer canceled his order, she can obtain an extra 10,000 yards of acetate. How many of each clothing item should TrendLines now produce to maximize profit? g. TrendLines assumes that it can sell every item that was not sold during September and October in a big sale in November at 60 percent of the original price. Therefore, it can sell all items in unlimited quantity during the November sale. (The previously mentioned upper limits on demand only concern the sales during September and October.) What should the new production plan be to maximize profit? Case 3-4 New Frontiers Rob Richman, president of AmeriBank, takes off his glasses, rubs his eyes in exhaustion, and squints at the clock in his study. It reads 3 AM. For the last several hours, Rob has been poring over ­AmeriBank’s financial statements from the last three quarters of operation. AmeriBank, a medium-sized bank with branches throughout the United States, is headed for dire economic straits. The bank, which provides transaction, savings, investment, and loan services, has been experiencing a steady decline in its net income over the past year, and trends show that the decline will continue. The bank is simply losing customers to nonbank and foreign bank competitors. AmeriBank is not alone in its struggle to stay out of the red. From his daily industry readings, Rob knows that many ­American banks have been suffering significant losses because of increasing competition from nonbank and foreign bank competitors offering services typically in the domain of American banks. Because the nonbank and foreign bank competitors specialize in particular services, they are able to better capture the market for those services by offering less expensive, more efficient, more convenient services. For example, large corporations now turn to foreign banks and commercial paper offerings for loans, and affluent Americans now turn to money-market funds for investment. Banks face the daunting challenge of distinguishing themselves from nonbank and foreign bank competitors. Rob has concluded that one strategy for distinguishing AmeriBank from its competitors is to improve services that ­ nonbank and foreign bank competitors do not readily provide: transaction services. He has decided that a more convenient transaction method must logically succeed the automatic teller machine, and he believes that electronic banking over the Internet allows this convenient transaction method. Over the Internet, customers are able to perform transactions on their desktop computers either at home or work. The explosion of the Internet means that most potential customers understand and use it. He therefore feels that if AmeriBank offers Web banking (as the practice of Internet banking is commonly called), the bank will attract many new customers. Before Rob undertakes the project to make Web banking possible, however, he needs to understand the market for Web banking and the services AmeriBank should provide over the Internet. For example, should the bank only allow customers to access account balances and historical transaction information over the Internet, or should the bank develop a strategy to allow customers to make deposits and withdrawals over the Internet? Should the bank try to recapture a portion of the investment market by continuously running stock prices and allowing customers to make stock transactions over the Internet for a minimal fee? Therefore, Rob has concluded that a major survey project should be undertaken to learn what customers want. Because AmeriBank is not in the business of performing surveys, Rob has decided to outsource the survey project to a professional survey company. He has opened the project up for bidding by several survey companies and will award the project Case 3-5 Assigning Students to Schools 119 to the company that is willing to perform the survey for the least cost. Rob provided each survey company with a list of survey requirements to ensure that AmeriBank receives the needed information for planning the Web banking project. Because different age groups require different services, AmeriBank is interested in surveying four different age groups. The first group encompasses customers who are 18 to 25 years old. The bank assumes that this age group has limited yearly income and performs minimal transactions. The second group encompasses customers who are 26 to 40 years old. This age group has significant sources of income, performs many transactions, requires numerous loans for new houses and cars, and invests in various securities. The third group encompasses customers who are 41 to 50 years old. These customers typically have the same level of income and perform the same number of transactions as the s­ econd age group, but the bank assumes that these customers are less likely to use Web banking since they have not become as comfortable with the explosion of computers or the Internet. Finally, the fourth group encompasses customers who are 51 years of age and over. These customers commonly crave security and require continuous information on retirement funds. The bank believes that it is highly unlikely that many customers in this age group will use Web banking, but the bank desires to learn the needs of this age group for the future. AmeriBank wants to interview 2,000 customers with at least 20 percent from the first age group, at least 27.5 percent from the second age group, at least 15 percent from the third age group, and at least 15 percent from the fourth age group. Rob understands that some customers are uncomfortable with using the Internet. He therefore wants to ensure that the survey includes a mix of customers who know the Internet well and those that have less exposure to the Internet. To ensure that AmeriBank obtains the correct mix, he wants to interview at least 15 percent of customers from the Silicon Valley where Internet use is high, at least 35 percent of customers from big cities where Internet use is medium, and at least 20 percent of customers from small towns where Internet use is low. Sophisticated Surveys is one of three survey companies competing for the project. It has performed an initial analysis of these survey requirements to determine the cost of surveying different populations. The costs per person surveyed are listed in the following table. Age Group Region Silicon Valley Big cities Small towns 18 to 25 26 to 40 41 to 50 51 and over $4.75 5.25 6.50 $6.50 5.75 7.50 $6.50 6.25 7.50 $5.00 6.25 7.25 Sophisticated Surveys explores the following options cumulatively. a. Formulate a linear programming model to minimize costs while meeting all survey constraints imposed by AmeriBank. b. If the profit margin for Sophisticated Surveys is 15 percent of cost, what bid will it submit? c. After submitting its bid, Sophisticated Surveys is informed that it has the lowest cost but that AmeriBank does not like the solution. Specifically, Rob feels that the selected survey population is not representative enough of the banking customer population. Rob wants at least 50 people of each age group surveyed in each region. What is the new bid made by Sophisticated Surveys? d. Rob feels that Sophisticated Surveys oversampled the 18-to25-year-old population and the Silicon Valley population. He imposes a new constraint that no more than 600 individuals can be surveyed from the 18-to-25-year-old population and no more than 650 individuals can be surveyed from the Silicon Valley population. What is the new bid? e. When Sophisticated Surveys calculated the cost of reaching and surveying particular individuals, the company thought that reaching individuals in young populations would be easiest. In a recently completed survey, however, Sophisticated Surveys learned that this assumption was wrong. The new costs for surveying the 18-to-25-year-old population are listed below. Region Silicon Valley Big cities Small towns Cost per Person $6.50 6.75 7.00 Given the new costs, what is the new bid? f. To ensure the desired sampling of individuals, Rob imposes even stricter requirements. He fixes the exact percentage of people that should be surveyed from each population. The requirements are listed next. Population 18 to 25 26 to 40 41 to 50 51 and over Silicon Valley Big cities Small towns Percentage of People Surveyed 25% 35 20 20 20 50 30 By how much would these new requirements increase the cost of surveying for Sophisticated Surveys? Given the 15 ­percent profit margin, what would Sophisticated Surveys bid? Case 3-5 Assigning Students to Schools The Springfield School Board has made the decision to close one of its middle schools (sixth, seventh, and eighth grades) at the end of this school year and reassign all of next year’s middle school students to the three remaining middle schools. The school district provides busing for all middle school students who must travel more than approximately a mile, so the school 120 Chapter Three Linear Programming: Formulation and Applications board wants a plan for reassigning the students that will minimize the total busing cost. The annual cost per student for busing from each of the six residential areas of the city to each of the schools is shown in the following table (along with other basic data for next year), where 0 indicates that busing is not needed and a dash indicates an infeasible assignment. How much does this increase the total busing cost? (This line of analysis will be pursued more rigorously in Case 7-3.) The school board is considering eliminating some busing to reduce costs. Option 1 is to only eliminate busing for students traveling 1 to 1.5 miles, where the cost per student is given in the table Busing Cost per Student Area Number of Students Percentage in 6th Grade Percentage in 7th Grade Percentage in 8th Grade School 1 School 2 School 3 450 600 550 350 500 450 32 37 30 28 39 34 38 28 32 40 34 28 30 35 38 32 27 38 $300 — 600 200 0 500 $ 0 400 300 500 — 300 $700 500 200 — 400 0 School capacity: 900 1,100 1,000 1 2 3 4 5 6 The school board also has imposed the restriction that each grade must constitute between 30 and 36 percent of each school’s population. The above table shows the percentage of each area’s middle school population for next year that falls into each of the three grades. The school attendance zone boundaries can be drawn so as to split any given area among more than one school, but assume that the percentages shown in the table will continue to hold for any partial assignment of an area to a school. You have been hired as a management science consultant to assist the school board in determining how many students in each area should be assigned to each school. a. Formulate and solve a linear programming model for this problem. b. What is your resulting recommendation to the school board? After seeing your recommendation, the school board expresses concern about all the splitting of residential areas among multiple schools. They indicate that they “would like to keep each neighborhood together.” c. Adjust your recommendation as well as you can to enable each area to be assigned to just one school. (Adding this restriction may force you to fudge on some other constraints.) as $200. Option 2 is to also eliminate busing for students traveling 1.5 to 2 miles, where the estimated cost per student is $300. d. Revise the model from part a to fit Option 1, and solve. Compare these results with those from part b, including the reduction in total busing cost. e. Repeat part d for Option 2. The school board now needs to choose among the three alternative busing plans (the current one or Option 1 or Option 2). One important factor is busing costs. However, the school board also wants to place equal weight on a second factor: the inconvenience and safety problems caused by forcing students to travel by foot or bicycle a substantial distance (more than a mile, and especially more than 1.5 miles). Therefore, they want to choose a plan that provides the best trade-off between these two factors. f. Use your results from parts b, d, and e to summarize the key information related to these two factors that the school board needs to make this decision. g. Which decision do you think should be made? Why? Note: This case will be continued in later chapters (Cases 5-4 and 7-3), so we suggest that you save your analysis, including your basic spreadsheet model. Case 3-6 Reclaiming Solid Wastes The Save-It Company operates a reclamation center that collects four types of solid waste materials and then treats them so that they can be amalgamated (treating and amalgamating are separate processes) into a salable product. Three different grades of this product can be made, depending on the mix of the materials used. (See the first table.) Although there is some flexibility in the mix for each grade, quality standards specify the minimum or maximum amount of the materials allowed in that product grade. (This minimum or maximum amount is the weight of the material expressed as a percentage of the total weight for that product grade.) For each of the two higher grades, a fixed percentage is specified for one of the materials. These specifications are given in the first table along with the cost of amalgamation and the selling price for each grade. The reclamation center collects its solid waste materials from some regular sources and so is normally able to maintain a steady rate for treating them. The second table gives the quantities available for collection and treatment each week, as well as the cost of treatment, for each type of material. Case 3-7 Project Pickings 121 Grade A Specification Material 1: Not more than 30% of the total Material 2: Not less than 40% of the total Material 3: Not more than 50% of the total Material 4: Exactly 20% of the total Amalgamation Cost per Pound Selling Price per Pound $3.00 $8.50 B Material 1: Not more than 50% of the total Material 2: Not less than 10% of the total Material 4: Exactly 10% of the total 2.50 7.00 C Material 1: Not more than 70% of the total 2.00 5.50 Material 1 2 3 4 Available Pounds/Week Treatment Cost per Pound 3,000 2,000 4,000 1,000 $3.00 6.00 4.00 5.00 The Save-It Company is solely owned by Green Earth, an organization that is devoted to dealing with environmental issues; Save-It’s profits are all used to help support Green Earth’s activities. Green Earth has raised contributions and grants, amounting to $30,000 per week, to be used exclusively to cover the entire treatment cost for the solid waste materials. The board of directors of Green Earth has instructed the management of Save-It to divide this money among the materials in such a way that at least half of the amount available of each material is actually collected and treated. These additional restrictions are listed in the second table. Within the restrictions specified in the two tables, management wants to allocate the materials to product grades so as to Additional Restrictions 1. For each material, at least half of the pounds/week available should be collected and treated. 2. $30,000 per week should be used to treat these materials. maximize the total weekly profit (total sales income minus total amalgamation cost). a. Formulate this problem in linear programming terms by identifying all the activities, resources, benefits, and fixed requirements that lurk within it. b. Formulate and solve a spreadsheet model for this linear programming problem. c. Express this linear programming model in the spreadsheet in algebraic form. Case 3-7 Project Pickings Tazer, a pharmaceutical manufacturing company, entered the pharmaceutical market 15 years ago with the introduction of six new drugs. Five of the six drugs were simply permutations of existing drugs and therefore did not sell very heavily. The sixth drug, however, addressed hypertension and was a huge success. Since Tazer had a patent on the hypertension drug, it experienced no competition, and profits from the hypertension drug alone kept Tazer in business. Pharmaceutical patents remain in force for 20 years, so this one has five more years before it expires. During the past 15 years, Tazer continued a moderate amount of research and development, but it never stumbled upon a drug as successful as the hypertension drug. One reason is that the company never had the motivation to invest heavily in innovative research and development. The company was riding the profit wave generated by its hypertension drug and did not feel the need to commit significant resources to finding new drug breakthroughs. Now Tazer is beginning to fear the pressure of competition. Tazer knows that once the patent expires in five years, generic drug manufacturing companies will swarm into the market like vultures. Historical trends show that generic drugs decrease sales of branded drugs by 75 percent. Tazer is therefore looking to invest significant amounts of money in research and development this year to begin the search for a new breakthrough drug that will offer the company the same success as the hypertension drug. Tazer believes that if the company begins extensive research and development now, the probability of finding a successful drug shortly after the expiration of the hypertension patent will be high. As head of research and development at Tazer, you are responsible for choosing potential projects and assigning project directors to lead each of the projects. After researching the needs of the market, analyzing the shortcomings of current drugs, 122 Chapter Three Linear Programming: Formulation and Applications and interviewing numerous scientists concerning the promising areas of medical research, you have decided that your department will pursue five separate projects, which are listed below: Project Up: Develop a more effective antidepressant that does not cause serious mood swings. Project Stable: Develop a drug that addresses manic-depression. Project Choice: Develop a less intrusive birth control method for women. Project Hope: Develop a vaccine to prevent HIV infection. Project Release: Develop a more effective drug to lower blood pressure. For each of the five projects, you are only able to specify the medical ailment the research should address since you do not know what compounds will exist and be effective without research. You also have five senior scientists to lead the five projects. You know that scientists are very temperamental people and will only work well if they are challenged and motivated by the project. To ensure that the senior scientists are assigned to projects they find motivating, you have established a bidding system for the projects. You have given each of the five scientists 1,000 bid points. They assign bids to each project, giving a higher number of bid points to projects they most prefer to lead. The following table provides the bids from the five senior scientists for the five individual projects. Project Project Up Project Stable Project Choice Project Hope Project Release Project Up: 20 Project Stable: 450 Project Choice: 451 Project Hope: 39 Project Release: 40 Under these new conditions with just four senior scientists, which scientists will lead which projects to maximize preferences? e. Do you support the assignments found in part d? Why or why not? f. Now you again consider all five scientists. You decide, however, that several scientists cannot lead certain projects. In particular, Dr. Mickey does not have experience with research on the immune system, so he cannot lead Project Hope. His family also has a history of manic-depression, and you feel that he would be too personally involved in Project Stable to serve as an effective project leader. Dr. Mickey therefore cannot lead Project Stable. Dr. Kvaal also does not have experience with research on the immune system and cannot lead Project Hope. In addition, Dr. Kvaal cannot lead Project Release because he does not have experience with research on the cardiovascular system. Finally, Dr. Rollins cannot lead Project Up because her family has a history of depression and you feel she would be too personally involved in the project to serve as an effective leader. Because Dr. Mickey and Dr. Kvaal cannot lead two of the five projects, they each have only 600 bid points. Dr. Rollins has only 800 bid points because she cannot lead Dr. Kvaal Dr. Zuner Dr. Tsai Dr. Mickey Dr. Rollins 100 400 200 200 100 0 200 800 0 0 100 100 100 100 600 267 153 99 451 30 100 33 33 34 800 You decide to evaluate a variety of scenarios you think are likely. a. Given the bids, you need to assign one senior scientist to each of the five projects to maximize the preferences of the scientists. What are the assignments? b. Dr. Rollins is being courted by Harvard Medical School to accept a teaching position. You are fighting desperately to keep her at Tazer, but the prestige of Harvard may lure her away. If this were to happen, the company would give up the project with the least enthusiasm. Which project would not be done? c. If Dr. Rollins leaves, you do not want to sacrifice any project since researching only four projects decreases the probability of finding a breakthrough new drug. You decide that either Dr. Zuner or Dr. Mickey could lead two projects. Under these new conditions with just four senior scientists, which scientists will lead which projects to maximize preferences? d. After Dr. Zuner was informed that she and Dr. Mickey are being considered for two projects, she decided to change her bids. Dr. Zuner’s new bids for each of the projects are shown next. one of the five projects. The following table provides the new bids of Dr. Mickey, Dr. Kvaal, and Dr. Rollins. Project Project Up Project Stable Project Choice Project Hope Project Release Dr. Mickey Dr. Kvaal Dr. Rollins 300 Can’t lead 125 Can’t lead 175 86 343 171 Can’t lead Can’t lead Can’t lead 50 50 100 600 Which scientists should lead which projects to maximize preferences? g. You decide that Project Hope and Project Release are too complex to be led by only one scientist. Therefore, each of these projects will be assigned two scientists as project leaders. You decide to hire two more scientists in order to staff all projects: Dr. Arriaga and Dr. Santos. Because of religious reasons, neither of them want to lead Project Choice and so they assign 0 bid points to this project. The next table lists all projects, scientists, and their bids. Case 3-7 Additional Cases 123 Project Dr. Kvaal Dr. Zuner Dr. Tsai Dr. Mickey Dr. Rollins Dr. Arriaga Dr. Santos Project Up Project Stable Project Choice Project Hope Project Release 86 343 171 Can’t lead Can’t lead 0 200 800 0 0 100 100 100 100 600 300 Can’t lead 125 Can’t lead 175 Can’t lead 50 50 100 600 250 250 0 250 250 111 1 0 333 555 Which scientists should lead which projects to maximize preferences? h. Do you think it is wise to base your decision in part g only on an optimal solution for a variant of an assignment problem? Additional Cases Additional cases for this chapter are also available at the University of Western Ontario Ivey School of Business website, cases. ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. Chapter Four The Art of Modeling with Spreadsheets Learning Objectives After completing this chapter, you should be able to 1. Describe the general process for modeling in spreadsheets. 2. Describe some guidelines for building good spreadsheet models. 3. Apply both the general process for modeling in spreadsheets and the guidelines in this chapter to develop your own spreadsheet model from a description of the problem. 4. Identify some deficiencies in a poorly formulated spreadsheet model. 5. Apply a variety of techniques for debugging a spreadsheet model. 124 Nearly all managers now make extensive use of spreadsheets to analyze business problems. What they are doing is modeling with spreadsheets. Spreadsheet modeling is a major emphasis throughout this book. Section 1.2 in Chapter 1 introduced a spreadsheet model for performing break-even analysis. Section 2.2 in ­Chapter 2 described how to use spreadsheets to formulate linear programming models. Chapter 3 focused on spreadsheet models for five key categories of linear programming problems: resource-allocation problems, cost-benefit-trade-off problems, mixed problems, transportation problems, and assignment problems. Many kinds of spreadsheet models are discussed in subsequent chapters as well. However, those presentations focus mostly on the characteristics of spreadsheet models that fit the management science techniques (such as linear programming) being covered in those chapters. We devote this chapter instead to the general process of building models with spreadsheets. Modeling in spreadsheets is more an art than a science. There is no systematic procedure that invariably will lead to a single correct spreadsheet model. For example, if two people were given exactly the same business problem to analyze with a spreadsheet, their spreadsheet models would likely look quite different. There is no one right way of modeling any given problem. However, some models will be better than others. Although no completely systematic procedure is available for modeling in spreadsheets, there is a general process that should be followed. This process has four major steps: (1) plan the spreadsheet model, (2) build the model, (3) test the model, and (4) analyze the model and its results. After introducing a case study in Section 4.1, the next section will describe this plan-build-test-analyze process in some detail and illustrate the process in the context of the case study. Section 4.2 also will discuss some ways of overcoming common stumbling blocks in the modeling process. Unfortunately, despite its logical approach, there is no guarantee that the plan-build-testanalyze process will lead to a “good” spreadsheet model. A good spreadsheet model is easy to understand, easy to debug, and easy to modify. Section 4.3 presents some guidelines for building such models. This section also uses the case study in Section 4.1 to illustrate the difference between appropriate formulations and poor formulations of a model. Even with an appropriate formulation, the initial versions of large spreadsheet models commonly will include some small but troublesome errors, such as inaccurate references to cell addresses or typographical errors when entering equations into cells. Indeed, some 4.1 A Case Study: The Everglade Golden Years Company Cash Flow Problem 125 studies have shown that a surprisingly large number of errors typically occur in the first version of such models. These errors often can be difficult to track down. Section 4.4 presents some helpful ways to debug a spreadsheet model and root out such errors. The overriding goal of this chapter is to provide a solid foundation for becoming a successful spreadsheet modeler. However, this chapter by itself will not turn you into a highly skilled modeler. Ultimately, to reach this point, you also will need to study various examples of good spreadsheet models in the different areas of management science and then have lots of practice in formulating your own models. This process will continue throughout the remainder of this book. 4.1 A CASE STUDY: THE EVERGLADE GOLDEN YEARS COMPANY CASH FLOW PROBLEM With only $1 million in cash reserves and negative cash flows looming soon, loans will be needed to satisfy the company policy of maintaining a balance of at least $500,000 at all times. The Everglade Golden Years Company operates upscale retirement communities in certain parts of southern Florida. The company was founded in 1946 by Alfred Lee, who was in the right place at the right time to enjoy many successful years during the boom in the Florida economy as many wealthy retirees flooded into the area. Today, the company continues to be run by the Lee family, with Alfred’s grandson, Sheldon Lee, as the CEO. The past few years have been difficult ones for Everglade. The demand for retirement community housing has been light and Everglade has been unable to maintain full occupancy. However, this market has picked up recently and the future is looking brighter. Everglade has recently broken ground for the construction of a new retirement community and has more new construction planned over the next 10 years (2018 through 2027). Julie Lee is the chief financial officer (CFO) at Everglade. She has spent the last week in front of her computer trying to come to grips with the company’s imminent cash flow problem. Julie has projected Everglade’s net cash flows over the next 10 years as shown in Table 4.1. With less money currently coming in than would be provided by full occupancy and with all the construction costs for the new retirement community, Everglade will have negative cash flow for the next few years. With only $1 million in cash reserves, it appears that Everglade will need to take out some loans in order to meet its financial obligations. Also, to protect against uncertainty, company policy dictates maintaining a balance of at least $500,000 in cash reserves at all times. The company’s bank has offered two types of loans to Everglade. The first is a 10-year loan with interest-only payments made annually and then the entire principal repaid in a single balloon payment after 10 years. The interest rate on this long-term loan is a favorable 5 percent per year. The second option is a series of one-year loans. These loans can be taken out each year as needed, but each must be repaid (with interest) the following year. Each new loan can be used to help repay the loan for the preceding year if needed. The interest rate for these short-term loans currently is projected to be 7 percent per year. Armed with her cash flow projections and the loan options from the bank, Julie schedules a meeting with the CEO, Sheldon Lee. Their discussion is as follows: Julie: Well, we really seem to be in a pickle. There is no way to meet our cash flow problems without borrowing money. TABLE 4.1 Projected Net Cash Flows for the Everglade Golden Years Company over the Next Ten Years Year Projected Net Cash Flow (millions of dollars) 2018 2019 2020 2021 2022 2023 2024 2025 2026 2027 −8 −2 −4 3 6 3 −4 7 −2 10 126 Chapter Four The Art of Modeling with Spreadsheets The objective is to develop a financial plan that will keep the company solvent and then maximize the cash balance in 2028, after all the loans are paid off. Sheldon: I was afraid of that. What are our options? Julie: I’ve talked to the bank, and we can take out a 10-year loan with an interest rate of 5 percent, or a series of one-year loans at a projected rate of 7 percent. Sheldon: Wow. That 5 percent rate sounds good. Can we just borrow all that we need using the 10-year loan? Julie: That was my initial reaction as well. However, after looking at the cash flow ­projections, I’m not sure the answer is so clear-cut. While we have negative cash flow for the next few years, the situation looks much brighter down the road. With a 10-year loan, we are obligated to keep the loan and make the interest payments for 10 years. The one-year loans are more flexible. We can borrow the money only in the years we need it. This way we can save on interest payments in the future. Sheldon: Okay. I can see how the flexibility of the one-year loans could save us some money. Those loans also will look better if interest rates come down in future years. Julie: Or they could go higher instead. There’s no way to predict future interest rates, so we might as well just plan on the basis of the current projection of 7 percent per year. Sheldon: Yes, you’re right. So which do you recommend, a 10-year loan or a series of one-year loans? Julie: Well, there’s actually another option as well. We could consider a combination of the two types of loans. We could borrow some money long-term to get the lower interest rate and borrow some money short-term to retain flexibility. Sheldon: That sounds complicated. What we want is a plan that will keep us solvent throughout the 10 years and then leave us with as large a cash balance as possible at the end of the 10 years after paying off all the loans. I also would like to see the amounts of the interest payments and when loan payments are due. Could you set this up on a spreadsheet to figure out the best plan? Julie: You bet. I’ll try that and get back to you. Sheldon: Great. Let’s plan to meet again next week when you have your report ready. You’ll see in the next two sections how Julie carefully develops her spreadsheet model for this cash flow problem. Review Questions 4.2 1. What is the advantage of the long-term loan for Everglade? 2. What is the advantage of the series of short-term loans for Everglade? 3. What is the objective for the financial plan that needs to be developed? OVERVIEW OF THE PROCESS OF MODELING WITH SPREADSHEETS Spaghetti code is a term from computer programming. It refers to computer code that is not logically organized and thus jumps all over the place, so it is jumbled like a plate of spaghetti. You will see later that a linear programming model can be incorporated into a spreadsheet to solve this problem. However, you also will see that the format of this spreadsheet model does not fit readily into any of the categories of linear programming models described in Chapter 3. Even the template given in Figure 3.8 that shows the format for a spreadsheet model of mixed problems (the broadest category of linear programming problems) does not help in formulating the model for the current problem. The reason is that the Everglade cash flow management problem is an example of a more complicated type of linear programming problem (a dynamic problem with many time periods) that requires starting from scratch in carefully formulating the spreadsheet model. Therefore, this example will nicely illustrate the process of modeling with spreadsheets when dealing with complicated problems of any type, including those discussed later in the book that do not fit linear programming. When presented with a problem like the Everglade problem, the temptation is to jump right in, launch Excel, and start entering a model. Resist this urge. Developing a spreadsheet model without proper planning inevitably leads to a model that is poorly organized and filled with “spaghetti code.” 4.2 Overview of the Process of Modeling with Spreadsheets 127 FIGURE 4.1 A flow diagram for the general plan-buildtest-analyze process for ­modeling with spreadsheets. Plan Build • Visualize where you want to ﬁnish • Do some calculations by hand • Sketch out a spreadsheet Start with a small-scale model Expand the model to full scale Test Try diﬀerent trial solutions to check the logic Analyze Evaluate proposed solutions and/or optimize with Solver If the solution reveals inadequacies in the model, return to Plan or Build A modeler might go back and forth between the Build and Test steps several times. Part of the challenge of planning and developing a spreadsheet model is that there is no standard procedure to follow. It is more an art than a science. However, to provide you with some structure as you begin learning this art, we suggest that you follow the modeling process depicted in Figure 4.1. As suggested by the figure, the four major steps in this process are to (1) plan, (2) build, (3) test, and (4) analyze the spreadsheet model. The process mainly flows in this order. However, the two-headed arrows between Build and Test indicate a back-and-forth process where testing frequently results in returning to the Build step to fix some problems discovered during the Test step. This back and forth movement between Build and Test may occur several times until the modeler is satisfied with the model. At the same time that this back and forth movement is occurring, the modeler may be involved with further building of the model. One strategy is to begin with a small version of the model to establish its basic logic and then, after testing verifies its accuracy, to expand to a full-scale model. Even after completing the testing and then the analyzing of the model, the process may return to the Build step or even the Plan step if the Analysis step reveals inadequacies in the model. Each of these four major steps may also include some detailed steps. For example, Figure 4.1 lists three detailed steps within the Plan step. Initially, when dealing with a fairly complicated problem, it is helpful to take some time to perform each of these detailed steps manually one at a time. However, as you become more experienced with modeling in spreadsheets, you may find yourself merging some of the detailed steps and quickly performing them mentally. An experienced modeler often is able to do some of these steps mentally, without working them out explicitly on paper. However, if you find yourself getting stuck, it is likely that you are missing a key element from one of the previous detailed steps. You then should go back a step or two and make sure that you have thoroughly completed those preceding steps. We now describe the various components of the modeling process in the context of the Everglade cash flow problem. At the same time, we also point out some common stumbling blocks encountered while building a spreadsheet model and how these can be overcome. Plan: Visualize Where You Want to Finish One common stumbling block in the modeling process occurs right at the very beginning. Given a complicated situation like the one facing Julie at Everglade, sometimes it can be difficult to decide how to even get started. At this point, rather than focusing on the model and how to get it started, it can be helpful to think about what you need when you are done. For example, in Julie’s case, she would be thinking now about what information she should 128 Chapter Four The Art of Modeling with Spreadsheets provide in her final report to Sheldon. What kinds of numbers need to be included in the report? The answer to this question can quickly lead you to the heart of the problem and help get the modeling process started. The main question Julie is faced with is which loan, or combination of loans, to use and in what amounts, so one set of numbers that are clearly needed will be the size of the loans to take. The long-term loan is taken in a single lump sum. Therefore, a single number is needed to indicate how much money to borrow now at the long-term rate. The short-term loan can be taken in any or all of the ten years, so there are ten numbers needed to indicate how much to borrow at the short-term rate in each given year. In the report, Julie would also need to assure Sheldon that the business will remain solvent through the next ten years. Therefore, the cash balances will need to be tracked over the next ten years to make sure they can maintain the needed minimum balance. The report would also include information on when loan and interest payments are due and in what amounts. Finally, any data that have been collected would be included, such as the various interest rates and projected net cash flows from the business. Table 4.2 summarizes the list of numbers that Julie would need to include in the final report. Given this list of numbers that are needed in the end, then you can start to think about how these numbers would fit into a model to try to find a good answer to the problem. Here are three questions you should ask yourself: • What are the decisions that need to be addressed in the model? (The numbers representing these decisions will need to be displayed as changing cells in the model.) • What is the objective (the overall measure of performance) that should be pursued when making these decisions? (The number representing the overall measure of performance will be the objective cell of the model.) • What are the other kinds of numbers that need to be included in the spreadsheet model? (Data will go into data cells. Certain other results will need to be calculated before making the decisions and these results will go into output cells.) The decisions that Julie needs to address are the size of the loans to take (the first two sets of numbers in Table 4.2). These 11 numbers will be the changing cells in the spreadsheet model. The objective when making these decisions is to maximize the cash balance after ten years while also keeping the company solvent through ten years. Therefore, the final cash balance will be the objective cell of the spreadsheet model. The other numbers in the list represent either data (projected net cash flow from the business each year and loan interest rates) or results (cash balances each year, loan payments each year, and interest payments each year). The data will go into data cells in the spreadsheet model, while the results will need to be calculated, and will go into output cells in the spreadsheet model. It is important to distinguish between the numbers that represent decisions (changing cells) and those that represent results (output cells). For instance, it may be tempting to include the cash balances as changing cells. These cells clearly change depending on the decisions made. However, the cash balances are a result of how much is borrowed, how much is paid, and all of the other cash flows. They cannot be chosen independently but instead are a function of the other numbers in the spreadsheet. The distinguishing characteristic of changing cells (the loan amounts) is that they do not depend on anything else. They represent the independent decisions being made. They impact the other numbers, but not vice versa. TABLE 4.2 List of numbers that Julie will need to include in her final report to Sheldon Numbers Needed for Final Report Size of long-term loan Size of short-term loans Cash balance each year Loan payments each year Interest payments each year Projected net cash flows each year Loan interest rates 4.2 Overview of the Process of Modeling with Spreadsheets 129 At this point, you should know what changing cells and output cells are needed. At this stage in the process, you should have a clear idea of what the answer will look like, including what and how many changing cells are needed, and what kind of results (output cells) should be obtained. Plan: Do Some Calculations by Hand When building a model, another common stumbling block can arise when trying to enter a formula in one of the output cells. For example, just how does Julie keep track of the cash balances in the Everglade cash flow problem? What formulas need to be entered? There are a lot of factors that enter into this calculation, so it is easy to get overwhelmed. If you are getting stuck at this point, it can be a very useful exercise to do some calculations by hand. Just pick some numbers for the changing cells and determine with a calculator or pencil and paper what the results should be. For example, pick some loan amounts for Everglade and then calculate the company’s resulting cash balance at the end of the first couple of years. To illustrate, suppose that Everglade takes a long-term loan of $6 million and then adds short-term loans of $2 million in 2018 and $5 million in 2019. How much cash would the company have left at the end of 2018 and at the end of 2019? These two quantities can be calculated by hand as follows. In 2018, Everglade has some initial money in the bank ($1 million), a negative cash flow from its business operations (− $8 million), and a cash inflow from the long-term and short-term loans ($6 million and $2 ­million, respectively). Thus, the ending balance for 2018 would be: Ending balance (2018) = Starting balance + Cash flow (2018) + LT loan (2018) + ST loan (2018) $1 million −$8 million +$6 million +$2 million $1 million The calculations for the year 2019 are a little more complicated. In addition to the starting balance left over from 2018 ($1 million), negative cash flow from business operations for 2019 (− $2 million), and a new short-term loan for 2019 ($5 million), the company will need to make interest payments on its 2018 loans as well as pay back the short-term loan from 2018. The ending balance for 2019 is therefore: Ending balance (2019) = Starting balance (from end of 2018) $1 million + Cash flow (2019) −$2 million + ST loan (2019) +$5 million −LT interest payment −(5%)($6 million) −ST interest payment −(7%)($2 million) −ST loan payback (2018) −$2 million $1.56 million Hand calculations can clarify what formulas are needed for the output cells. Doing calculations by hand can help in a couple of ways. First, it can help clarify what formula should be entered for an output cell. For instance, looking at the by-hand calculations above, it appears that the formula for the ending balance for a particular year should be Ending balance = Starting balance + Cash flow + Loans − Interest payments − Loan paybacks Once a spreadsheet is laid out, this equation will make it straightforward to enter the proper cell references in the formula for the ending balance for any year in the spreadsheet model. The second advantage of using hand calculations is that they can help to verify the spreadsheet model after it has been constructed. By plugging in a long-term loan of $6 million, along with short-term loans of $2 million in 2018 and $5 million in 2019, into a completed spreadsheet, the ending balances should be the same as calculated above. If they’re not, this suggests an error in the spreadsheet model (assuming the hand calculations are correct). 130 Chapter Four The Art of Modeling with Spreadsheets Plan: Sketch Out a Spreadsheet Plan where the various blocks of data cells, changing cells, and output cells should go on the spreadsheet by sketching your layout ideas on paper. Any model typically has a large number of different elements that need to be included on the spreadsheet. For the Everglade problem, these would include some data cells (interest rates, starting balance, minimum balances, and cash flows), some changing cells (loan amounts), and a number of output cells (interest payments, loan paybacks, and ending balances). Therefore, a potential stumbling block can arise when trying to organize and lay out the spreadsheet model. Where should all the pieces fit on the spreadsheet? How do you begin putting together the spreadsheet? Before firing up Excel and blindly entering the various elements, it can be helpful to sketch a layout of the spreadsheet. Is there a logical way to arrange the elements? A little planning at this stage can go a long way toward building a spreadsheet that is well organized. Don’t bother with numbers at this point. Simply sketch out blocks on a piece of paper for the various data cells, changing cells, and output cells, and label them. Concentrate on the layout. When sketching the layout, a key question for each block of numbers is whether it should be laid out in a row or a column, or as a two-dimensional table? When addressing this question for the various blocks of numbers, you should also ask whether there are common row or column headings for different blocks of cells? If so, try to arrange the blocks in consistent rows or columns so they can utilize a single set of headings. Another guideline for sketching the layout is to try to arrange the spreadsheet so that it starts with the data at the top and progresses logically toward the objective cell at the bottom. This will be easier to understand and follow than if the data cells, changing cells, output cells, and objective cell are all scattered throughout the spreadsheet. A sketch of a potential spreadsheet layout for the Everglade problem is shown in Figure 4.2. The data cells for the interest rates, starting balance, and minimum cash balance are at the top of the spreadsheet. All of the remaining elements in the spreadsheet then follow the same structure. The rows represent the different years (from 2018 through 2028). All the various cash inflows and outflows are then broken out in the columns, starting with the projected cash flow from the business operations (with data for each of the 10 years), continuing with the loan inflows, interest payments, and loan paybacks, and culminating with the ending balance (calculated for each year). The long-term loan is a one-time loan (in 2018), so it is sketched as a single cell. The short-term loan can occur in any of the 10 years (2018 through 2027), so it is sketched as a block of cells. The interest payments start one year after the loans. The longterm loan is paid back 10 years later (2028). Organizing the elements with a consistent structure, like in Figure 4.2, not only saves having to retype the year labels for each element, but also makes the model easier to understand. Everything that happens in a given year is arranged together in a single row. FIGURE 4.2 Sketch of the spreadsheet for Everglade’s cash flow problem. LT Rate ST Rate Start Balance Minimum Cash Cash Flow LT Loan ST Loan LT Interest ST Interest LT Payback ST Payback Ending Balance Minimum Balance 2018 2019 . . . . 2027 2028 ≥ 4.2 Overview of the Process of Modeling with Spreadsheets 131 It is generally easiest to start sketching the portion of the layout where the data would be placed. The structure of the rest of the model should then follow the structure of the data cells. For example, once the projected cash flows data are sketched as a vertical column (with each year in a row), then it follows that the other cash flows should be structured the same way. Note that the sketch of the layout shown in Figure 4.2 has a logical progression to the spreadsheet. The data for the problem are located at the top and left of the spreadsheet. Then, since the cash flow, loan amounts, interest payments, and loan paybacks are all part of the calculation for the ending balance, the columns are arranged this way, with the ending balance directly to the right of all these other elements. Since Sheldon has indicated that the objective is to maximize the ending balance in 2028, this cell is designated to be the objective cell. Each year, the balance must be greater than the minimum required balance ($500,000). Since this will be a constraint in the model, it is logical to arrange the balance and minimum balance blocks of numbers adjacent to each other in the spreadsheet. You can put the ≥ signs on the sketch to remind yourself that these will be constraints. The sketch of a spreadsheet in Figure 4.2 has a logical progression, starting with the data on the top left and then moving through the calculations toward the objective cell on the bottom right. Build: Start with a Small Version of the Spreadsheet Once you’ve thought about a logical layout for the spreadsheet, it is finally time to open a new worksheet in Excel and start building the model. If it is a complicated model, you may want to start by building a small, readily manageable version of the model. The idea is to first make sure that you’ve got the logic of the model worked out correctly for the small version before expanding the model to full scale. For example, in the Everglade problem, we could get started by building a model for just the first two years (2018 and 2019), like the spreadsheet shown in Figure 4.3. Work out the logic for a small version of the spreadsheet model before expanding to full size. FIGURE 4.3 A small version (years 2018 and 2019 only) of the spreadsheet for the Everglade cash flow management problem. A 1 B C D E F G I J ST Payback Ending Balance H K L Everglade Cash Flow Management Problem (Years 2018 and 2019) 2 3 LT Rate 5% 4 ST Rate 7% (all cash figures in millions of dollars) 5 6 Start Balance 1 7 Minimum Cash 0.5 10 Year Cash Flow LT Loan ST Loan 11 2018 –8 6 2 12 2019 –2 8 9 LT Interest ST Interest –0.30 –0.14 5 LT Payback –2.00 F G H I J 9 LT ST LT ST 10 Interest Interest Payback Payback Ending Balance = –LTRate*LTLoan = –STRate*E11 11 12 Range Name Cell LTLoan LTRate MinimumCash StartBalance STRate D11 C3 C7 C6 C4 = –E11 Minimum Balance 1.00 ≥ 0.50 1.56 ≥ 0.50 =StartBalance+SUM(C11:I11) =J11+SUM(C12:I12) K L Minimum Balance ≥ ≥ =MinimumCash =MinimumCash 132 Chapter Four The Art of Modeling with Spreadsheets This spreadsheet is set up to follow the layout suggested in the sketch of Figure 4.2. The loan amounts are in columns D and E. Since the interest payments are not due until the following year, the formulas in columns F and G refer to the loan amounts from the preceding year (LTLoan, or D11, for the long-term loan, and E11 for the short-term loan). The loan payments are calculated in columns H and I. Column H is blank because the long-term loan does not need to be repaid until 2028. The short-term loan is repaid one year later, so the formula in cell I12 refers to the short-term loan taken the preceding year (cell E11). The ending balance in 2018 is the starting balance plus the sum of all the various cash flows that occur in 2018 (cells C11:I11). The ending balance in 2019 is the ending balance in 2018 (cell J11) plus the sum of all the various cash flows that occur in 2019 (cells C12:I12). All these formulas are summarized below the spreadsheet in Figure 4.3. Building a small version of the spreadsheet works very well for spreadsheets that have a time dimension. For example, instead of jumping right into a 10-year planning problem, we can start with the simpler problem of just looking at a couple of years. Once this smaller model is working correctly, you then can expand the model to 10 years. Even if a spreadsheet model does not have a time dimension, the same concept of starting small can be applied. For example, if certain constraints considerably complicate a problem, start by working on a simpler problem without the difficult constraints. Get the simple model working and then move on to tackle the difficult constraints. If a model has many sets of output cells, you can build up a model piece by piece by working on one set of output cells at a time, making sure each set works correctly before moving on to the next. Test: Test the Small Version of the Model Try entering numbers in the changing cells for which you know what the values of the output cells should be. If you do start with a small version of the model first, be sure to test this version thoroughly to make sure that all the logic is correct. It is far better to fix a problem early, while the spreadsheet is still a manageable size, rather than later after an error has been propagated throughout a much larger spreadsheet. To test the spreadsheet, try entering values in the changing cells for which you know what the values of the output cells should be, and then see if the spreadsheet gives the results that you expect. For example, in Figure 4.3, if zeroes are entered for the loan amounts, then the interest payments and loan payback quantities also should be zero. If $1 million is borrowed for both the long-term loan and the short-term loan, then the interest payments the following year should be $50,000 and $70,000, respectively. (Recall that the interest rates are 5 percent and 7 percent, respectively.) If Everglade takes out a $6 million long-term loan and a $2 million short-term loan in 2018, plus a $5 million short-term loan in 2019, then the ending balances should be $1 million for 2018 and $1.56 million for 2019 (based on the calculations done earlier by hand). All these tests work correctly for the spreadsheet in Figure 4.3, so we can be fairly certain that it is correct. If the output cells are not giving the results that you expect, then carefully look through the formulas to see if you can determine and fix the problem. Section 4.4 will give further guidance on some ways to debug a spreadsheet model. Build: Expand the Model to Full-Scale Size Excel Tip: A shortcut for filling down or filling across to the right is to select the cell you want to copy, click on the fill handle (the small box in the lower right-hand corner of the selection rectangle), and drag through the cells you want to fill. Once a small version of the spreadsheet has been tested to make sure all the formulas are correct and everything is working properly, the model can be expanded to full-scale size. Excel’s fill commands often can be used to quickly copy the formulas into the remainder of the model. To illustrate this process, consider the small version of the Everglade spreadsheet in Figure 4.3, The formulas in columns F, G, I, J, and L can be copied using the Fill Down command in the Editing Group of the Home tab to obtain all the formulas shown in Figure 4.4. For example, selecting cells G12:G21 and choosing Fill Down will take the formula in cell G12 and copy it (while adjusting the cell address in Column E for the formula) into cells G13 through G21. We explain next how this works. Before using the fill commands to copy formulas, be sure that the relative and absolute ­references have been used appropriately. (Appendix A provides details about relative and absolute references.) A reference to a range name is an absolute reference. For example, in G12 (= − STRate*E11), using a range name STRate for the short-term loan interest rate 4.2 Overview of the Process of Modeling with Spreadsheets 133 FIGURE 4.4 A complete spreadsheet model for the Everglade cash flow management problem, including the equations entered into the objective cell EndBalance (J21) and all the other output cells, to be used before calling on Solver. The entries in the changing cells, LTLoan (D11) and STLoan (E11:E20), are only a trial solution at this stage. A 1 B C E D F G H I J K L Everglade Cash Flow Management Problem 2 3 LT Rate 5% 4 ST Rate 7% (all cash figures in millions of dollars) 5 6 Start Balance 1 7 Minimum Cash 0.5 Cash LT ST LT ST LT ST Ending 10 Year Flow Loan Loan Interest Interest Payback Payback Balance 11 2018 –8 6 2 8 9 Minimum Balance 1.00 ≥ 0.5 1.56 ≥ 0.5 12 2019 –2 5 –0.30 –0.14 –2 13 2020 –4 0 –0.30 –0.35 –5 –8.09 ≥ 0.5 14 2021 3 0 –0.30 0 0 –5.39 ≥ 0.5 15 2022 6 0 –0.30 0 0 0.31 ≥ 0.5 16 2023 3 0 –0.30 0 0 3.01 ≥ 0.5 17 2024 –4 0 –0.30 0 0 –1.29 ≥ 0.5 18 2025 7 0 –0.30 0 0 5.41 ≥ 0.5 19 2026 –2 0 0 3.11 ≥ 0.5 2027 10 0 –0.30 –0.30 0 20 0 0 12.81 ≥ 0.5 21 2028 –0.30 0 0 6.51 ≥ 0.5 F –6 G H I J 9 LT ST LT ST Ending 10 Interest Interest Payback Payback 11 Balance K L Minimum Balance =StartBalance+SUM(C11:I11) ≥ =MinimumCash 12 =–LTRate*LTLoan =–STRate*E11 =–E11 =J11+SUM(C12:I12) ≥ =MinimumCash 13 =–LTRate*LTLoan =–STRate*E12 =–E12 =J12+SUM(C13:I13) ≥ =MinimumCash 14 =–LTRate*LTLoan =–STRate*E13 =–E13 =J13+SUM(C14:I14) ≥ =MinimumCash 15 =–LTRate*LTLoan =–STRate*E14 =–E14 =J14+SUM(C15:I15) ≥ =MinimumCash 16 =–LTRate*LTLoan =–STRate*E15 =–E15 =J15+SUM(C16:I16) ≥ =MinimumCash 17 =–LTRate*LTLoan =–STRate*E16 =–E16 =J16+SUM(C17:I17) ≥ =MinimumCash 18 =–LTRate*LTLoan =–STRate*E17 =–E17 =J17+SUM(C18:I18) ≥ =MinimumCash 19 =–STRate*E18 =–E18 =J18+SUM(C19:I19) ≥ =MinimumCash 20 =–LTRate*LTLoan =–LTRate*LTLoan =–STRate*E19 =–E19 =J19+SUM(C20:I20) ≥ =MinimumCash 21 =–LTRate*LTLoan =–STRate*E20 =–LTLoan =–E20 =J20+SUM(C21:I21) ≥ =MinimumCash Range Name Cells CashFlow EndBalance EndingBalance LTLoan LTRate MinimumBalance MinimumCash StartBalance STLoan STRate C11:C20 J21 J11:J21 D11 C3 L11:L21 C7 C6 E11:E20 C4 134 Chapter Four The Art of Modeling with Spreadsheets Excel Tip: A shortcut for changing a cell reference from relative to absolute is to press the F4 key on a PC or command-T on a Mac. makes this an absolute reference. When copied into cells G13:G21, the short-term loan interest rate used will always be the value in STRate (C4). By contrast, a reference to a specific cell address is a relative reference. For example, the reference to E11 (the loan amount from the previous year) in the formula for G12 is a relative reference. E11 is two cells to the left and one cell up. When the formula is copied from G12 into G13:G21, the reference in each of these cells will continue to be two cells to the left and one cell up. This is exactly what we want, since we always want the interest payment to be based on the short-term loan that was taken one year ago (two cells to the left and one cell up). After using the Fill Down command to copy the formulas in columns F, G, I, J, and L and entering the LT loan payback into cell H21, the complete model appears as shown in Figure 4.4. Test: Test the Full-Scale Version of the Model Just as it was important to test the small version of the model, it needs to be tested again after it is expanded to full-scale size. The procedure is the same one followed for testing the small version, including the ideas that will be presented in Section 4.4 for debugging a spreadsheet model. Analyze: Analyze the Model Before using Solver, the spreadsheet in Figure 4.4 is merely an evaluative model for Everglade. It can be used to evaluate any proposed solution, including quickly determining what interest and loan payments will be required and what the resulting balances will be at the end of each year. For example, LTLoan (D11) and STLoan (E11:E20) in Figure 4.4 show one ­possible plan, which turns out to be unacceptable because EndingBalance (J11:J21) indicates that a negative ending balance would result in three of the years. To optimize the model, Solver is used as shown in Figure 4.5 to specify the objective cell, the changing cells, and the constraints. Everglade management wants to find a combination of loans that will keep the company solvent throughout the next 10 years (2018–2027) and then will leave as large a cash balance as possible in 2028 after paying off all the loans. Therefore, the objective cell to be maximized is EndBalance (J21) and the changing cells are the loan amounts LTLoan (D11) and STLoan (E11:E20). To assure that Everglade maintains a minimum balance of at least $500,000 at the end of each year, the constraints for the model are EndingBalance (J11:J21) ≥ MinimumBalance (L11:L21). After running Solver, the optimal solution is shown in Figure 4.5. The changing cells, LTLoan (D11) and STLoan (E11:E20), give the loan amounts in the various years. The objective cell EndBalance (J21) indicates that the ending balance in 2028 will be $5.39 million. Conclusion of the Case Study The spreadsheet model developed by Everglade’s CFO, Julie Lee, is the one shown in ­Figure 4.5. Her next step is to submit a report to her CEO, Sheldon Lee, that recommends the plan obtained by this model. Soon thereafter, Sheldon and Julie meet to discuss her report. Sheldon: Thanks for your report, Julie. Excellent job. Your spreadsheet really lays everything out in a very understandable way. Julie: Thanks. It took a little while to get the spreadsheet organized properly and to make sure it was operating correctly, but I think the time spent was worthwhile. Sheldon: Yes, it was. You can’t rush those things. But one thing is still bothering me. Julie: What’s that? Sheldon: It has to do with our forecasts for the company’s future cash flows. We have been assuming that the cash flows in the coming years will be the ones shown in column C of your spreadsheet. Those are good estimates, but we both know that they are only estimates. A lot of changes that we can’t foresee now are likely to occur over the next 10 years. When there is a shift in the economy, or when other unexpected developments occur that impact the company, those cash flows can change a lot. How do we know if your recommended plan will still be a good one if those kinds of changes occur? 4.2 Overview of the Process of Modeling with Spreadsheets 135 FIGURE 4.5 A complete spreadsheet model for the Everglade cash flow management problem after calling on Solver to obtain the optimal solution shown in the changing cells, LTLoan (D11) and STLoan (E11:E20). The objective cell EndBalance (J21) indicates that the resulting cash balance in 2028 will be $5.39 million if all the data cells prove to be accurate. A 1 B C E D F G H I J K L Everglade Cash Flow Management Problem 2 3 LT Rate 5% 4 ST Rate 7% (all cash figures in millions of dollars) 5 6 Start Balance 1 7 Minimum Cash 0.5 Cash LT ST LT ST LT ST Ending 10 Year Flow Loan Loan Interest Interest Payback Payback Balance 11 2018 –8 4.65 2.85 12 2019 –2 5.28 –0.23 –0.20 13 2020 –4 9.88 –0.23 14 2021 3 7.81 –0.23 15 2022 6 2.59 –0.23 16 2023 3 0 17 2024 –4 18 2025 19 8 9 Minimum Balance 0.50 ≥ –2.85 0.50 ≥ 0.50 –0.37 –5.28 0.50 –9.88 0.50 –0.55 –7.81 0.50 ≥ ≥ ≥ 0.50 –0.69 –0.23 –0.18 –2.59 0.50 ≥ 0.50 4.23 –0.23 0 0 0.50 ≥ 0.50 7 0 –0.23 –0.30 – 4.23 2.74 ≥ 0.50 2026 –2 0 0 0 0.51 2027 10 0 0 0 10.27 ≥ ≥ 0.50 20 –0.23 –0.23 21 2028 –0.23 0 0 5.39 ≥ 0.50 –4.65 F G H I J 9 LT ST LT ST Ending 10 Interest Interest Payback Payback 11 0.50 0.50 0.50 0.50 K L Minimum Balance Balance = StartBalance+SUM(C11:I11) ≥ =MinimumCash 12 =–LTRate*LTLoan =–STRate*E11 =–E11 =J11+SUM(C12:I12) ≥ =MinimumCash 13 =–LTRate*LTLoan =–STRate*E12 =–E12 =J12+SUM(C13:I13) 14 =–LTRate*LTLoan =–STRate*E13 =–E13 =J13+SUM(C14:I14) 15 =–LTRate*LTLoan =–STRate*E14 =–E14 =J14+SUM(C15:I15) ≥ =MinimumCash ≥ =MinimumCash ≥ =MinimumCash 16 =–LTRate*LTLoan =–STRate*E15 =–E15 =J15+SUM(C16:I16) ≥ =MinimumCash 17 =–LTRate*LTLoan =–STRate*E16 =–E16 =J16+SUM(C17:I17) ≥ =MinimumCash 18 =–LTRate*LTLoan =–STRate*E17 =–E17 =J17+SUM(C18:I18) 19 ≥ =MinimumCash =–STRate*E18 =–E18 =J18+SUM(C19:I19) 20 =–LTRate*LTLoan =–LTRate*LTLoan =–STRate*E19 =–E19 =J19+SUM(C20:I20) ≥ =MinimumCash ≥ =MinimumCash 21 =–LTRate*LTLoan =–STRate*E20 =–LTLoan =–E20 =J20+SUM(C21:I21) ≥ =MinimumCash Solver Parameters Set Objective Cell: EndBalance To: Max By Changing Variable Cells: LTLoan, STLoan Subject to the Constraints: EndingBalance >= MinimumBalance Solver Options: Make Variables Nonnegative Solving Method: Simplex LP Range Name Cells CashFlow EndBalance EndingBalance LTLoan LTRate MinimumBalance MinimumCash StartBalance STLoan STRate C11:C20 J21 J11:J21 D11 C3 L11:L21 C7 C6 E11:E20 C4 136 Chapter Four The Art of Modeling with Spreadsheets Julie: A very good question. To answer it, we should do some what-if analysis to see what would happen if those kinds of changes occur. Now that the spreadsheet is set up properly, it will be easy to do that by simply changing some of the cash flows in column C and seeing what would happen with the current plan. You can try out any change or changes you want and immediately see the effect. Each time you change a future cash flow, you also have the option of trying out changes on short-term loan amounts to see what kind of adjustments would be needed to maintain a balance of at least $500,000 in every year. OK, are you ready? Shall we do some what-if analysis now? Sheldon: Let’s do. Providing a data cell for each piece of data makes it easy to check what would happen if the correct value for a piece of data differs from its initial estimate. Fortunately, Julie had set up the spreadsheet properly (providing a data cell for the cash flow in each of the next 10 years) to enable performing what-if analysis immediately by simply trying different numbers in some of these data cells. (The next chapter will focus on describing the importance of what-if analysis and alternative ways of performing this kind of analysis.) After spending half an hour trying different numbers, Sheldon and Julie conclude that the plan in Figure 4.5 will be a sound initial financial plan for the next 10 years, even if future cash flows deviate somewhat from current forecasts. If deviations do occur, adjustments will of course need to be made in the short-term loan amounts. At any point, Julie also will have the option of returning to the company’s bank to try to arrange another long-term loan for the remainder of the 10 years at a lower interest rate than that offered for short-term loans. If so, essentially the same spreadsheet model as in Figure 4.5 can be used, along with Solver, to find the optimal adjusted financial plan for the remainder of the 10 years. A management science technique called computer simulation provides another effective way of taking the uncertainty of future cash flows into account. Chapters 12 and 13 will describe this technique and then Problem 13.24 will continue the analysis of this same case study. Review Questions 1. What is a good way to get started with a spreadsheet model if you don’t even know where to begin? 2. What are two ways in which doing calculations by hand can help you? 3. Describe a useful way to get started organizing and laying out a spreadsheet. 4. What types of values should be put into the changing cells to test the model? 5. What is the difference between an absolute cell reference and a relative cell reference? 4.3 SOME GUIDELINES FOR BUILDING “GOOD” SPREADSHEET MODELS There are many ways to set up a model on a spreadsheet. While one of the benefits of spreadsheets is the flexibility they offer, this flexibility also can be dangerous. Although Excel provides many features (such as range names, shading, borders, etc.) that allow you to create “good” spreadsheet models that are easy to understand, easy to debug, and easy to modify, it is also easy to create “bad” spreadsheet models that are difficult to understand, difficult to debug, and difficult to modify. The goal of this section is to provide some guidelines that will help you to create “good” spreadsheet models. Enter the Data First All the data should be laid out on the spreadsheet before beginning to formulate the rest of the spreadsheet model. Any spreadsheet model is driven by the data in the spreadsheet. The form of the entire model is built around the structure of the data. Therefore, it is always a good idea to enter and carefully lay out all the data before you begin to set up the rest of the model. The model structure then can conform to the layout of the data as closely as possible. The reason for entering the data first is that it is easier to set up the rest of the model when the data are already on the spreadsheet. In the Everglade problem (see Figure 4.5), the data for the cash flows have been laid out in the first columns of the spreadsheet (B and C), with the year labels in column B and the data in cells C11:C20. Once the data are in place, the layout for the rest of the model quickly falls into place around the structure of the data. It is only logical to lay out the changing cells and output cells using the same structure, with each of the various cash flows in columns that utilize the same row labels from column B. An Application Vignette Welch’s, Inc., is the world’s largest processor of Concord and Niagara grapes with large annual sales that reached $608 million in 2013. Such products as Welch’s grape jelly and Welch’s grape juice have been enjoyed by generations of American consumers. Every September, growers begin delivering grapes to processing plants that then press the raw grapes into juice. Time must pass before the grape juice is ready for conversion into finished jams, jellies, juices, and concentrates. Deciding how to use the grape crop is a complex task given changing demand and uncertain crop quality and quantity. Typical decisions include what recipes to use for major product groups, the transfer of grape juice between plants, and the mode of transportation for these transfers. Because Welch’s lacked a formal system for optimizing raw material movement and the recipes used for production, a management science team developed a preliminary linear programming model. This was a large model with 8,000 decision variables that focused on the component level of detail. Smallscale testing proved that the model worked. To make the model more useful, the team then revised it by aggregating demand by product group rather than by component. This reduced its size to 324 decision variables and 361 functional constraints. The model then was incorporated into a spreadsheet. The company has run the continually updated version of this spreadsheet model each month to provide senior management with information on the optimal logistics plan generated by Solver. The savings from using and optimizing this model were approximately $150,000 in the first year alone. A major advantage of incorporating the linear programming model into a spreadsheet has been the ease of explaining the model to managers with differing levels of mathematical understanding. This has led to a widespread appreciation of the management science approach for both this application and others. Source: E. W. Schuster and S. J. Allen, “Raw Material Management at Welch’s, Inc.,” Interfaces 28, no. 5 (September—October 1998), pp. 13–24. (A link to this article is available at www.mhhe.com/Hillier6e.) Now reconsider the spreadsheet model developed in Section 2.2 for the Wyndor Glass Co. problem. The spreadsheet is repeated here in Figure 4.6. The data for the Hours Used per Unit Produced have been laid out in the center of the spreadsheet in cells C7:D9. The output cells, HoursUsed (E7:E9), then have been placed immediately to the right of these data and to the left of the data on HoursAvailable(G7:G9), where the row labels for these output cells are the same as for all these data. This makes it easy to interpret the three constraints being laid out in rows 7–9 of the spreadsheet model. Next, the changing cells and objective cell have been placed together in row 12 below the data, where the column labels for the changing cells are the same as for the columns of data above. The locations of the data occasionally will need to be shifted somewhat to better accommodate the overall model. However, with this caveat, the model structure generally should conform to the data as closely as possible. Organize and Clearly Identify the Data Provide labels in the spreadsheet that clearly identify all the data. Related data should be grouped together in a convenient format and entered into the spreadsheet with labels that clearly identify the data. For data that are laid out in tabular form, the table should have a heading that provides a general description of the data and then each row and column should have a label that will identify each entry in the table. The units of the data also should be identified. Different types of data should be well separated in the spreadsheet. However, if two tables need to use the same labels for either their rows or columns, then be consistent in making them either rows in both tables or columns in both tables. In the Wyndor Glass Co. problem (Figure 4.6), the three sets of data have been grouped into tables and clearly labeled Unit Profit, Hours Used per Unit Produced, and Hours Available. The units of the data are identified (dollar signs are included in the unit profit data and hours are indicated in the labels of the time data). Finally, all three data tables make consistent use of rows and columns. Since the Unit Profit data have their product labels (Doors and Windows) in columns C and D, the Hours Used per Unit Produced data use this same structure. This structure also is carried through to the changing cells (Units Produced). Similarly, the data for each plant (rows 7–9) are in the rows for both the Hours Used per Unit ­Produced data and the Hours Available data. Keeping the data oriented the same way is not only less confusing, but it also makes it possible to use the SUMPRODUCT function. Recall that the SUMPRODUCT function assumes that the two ranges are exactly the same shape (i.e., the same number of rows and the same number of columns). If the Unit Profit data and the Units 137 138 Chapter Four The Art of Modeling with Spreadsheets FIGURE 4.6 The spreadsheet model formulated in Section 2.2 for the Wyndor Glass Co. product-mix problem. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Used Hours Available 7 Plant 1 1 0 2 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 2 6 $3,600 10 11 12 Units Produced Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: UnitsProduced Subject to the Constraints: HoursUsed <= HoursAvailable Solver Options: Make Variables Nonnegative Solving Method: Simplex LP E 5 Hours 6 7 8 9 Used =SUMPRODUCT(C7:D7,UnitsProduced) =SUMPRODUCT(C8:D8,UnitsProduced) =SUMPRODUCT(C9:D9,UnitsProduced) G 11 12 Total Profit =SUMPRODUCT(Unit Profit,UnitsProduced) Range Name Cells HoursAvailable HoursUsed HoursUsedPerUnitProduced TotalProfit UnitProfit UnitsProduced G7:G9 E7:E9 C7:D9 G12 C4:D4 C12:D12 Produced data had not been oriented the same way (e.g., one in a column and the other in a row), it would not have been possible to use the SUMPRODUCT function in the Total Profit calculation. Similarly, for the Everglade problem (Figure 4.5), the five sets of data have been grouped into cells and tables and clearly labeled ST Rate, LT Rate, Start Balance, Cash Flow, and Minimum Cash. The units of the data are identified (cells F5:I5 specify that all cash figures are in millions of dollars), and all the tables make consistent use of rows and columns (years in the rows). Enter Each Piece of Data into One Cell Only Every formula using the same piece of data should refer to the same single data cell. If a piece of data is needed in more than one formula, then refer to the original data cell rather than repeating the data in additional places. This makes the model much easier to modify. If the value of that piece of data changes, it only needs to be changed in one place. You do not need to search through the entire model to find all the places where the data value appears. For example, in the Everglade problem (Figure 4.5), there is a company policy of maintaining a cash balance of at least $500,000 at all times. This translates into a constraint for the minimum balance of $500,000 at the end of each year. Rather than entering the minimum cash position of 0.5 (in millions of dollars) into all the cells in column L, it is entered once in MinimumCash (C7) and then referred to by the cells in MinimumBalance (L11:L21). Then, if this policy were to change to, say, a minimum of $200,000 cash, the number would need to be changed in only one place. 4.3 Some Guidelines for Building “Good” Spreadsheet Models 139 Separate Data from Formulas Formulas should refer to data cells for any needed numbers. Avoid using numbers directly in formulas. Instead, enter any needed numbers into data cells and then refer to the data cells as needed. For example, in the Everglade problem (Figure 4.5), all the data (the interest rates, starting balance, minimum cash, and projected cash flows) are entered into separate data cells on the spreadsheet. When these numbers are needed to calculate the interest charges (in columns F and G), loan payments (in column H and I), ending balances (column J), and minimum balances (column L), the data cells are referred to rather than entering these numbers directly in the formulas. Separating the data from the formulas has a couple advantages. First, all the data are visible on the spreadsheet rather than buried in formulas. Seeing all the data makes the model easier to interpret. Second, the model is easier to modify since changing data only requires modifying the corresponding data cells. You don’t need to modify any formulas. This proves to be very important when it comes time to perform what-if analysis (the subject of the next chapter) to see what the effect would be if some of the estimates in the data cells were to take on other plausible values. Keep It Simple Make the spreadsheet as easy to interpret as possible. Avoid the use of powerful Excel functions when simpler functions are available that are easier to interpret. As much as possible, stick to SUMPRODUCT or SUM functions. This makes the model easier to understand and also helps to ensure that the model will be linear. (Linear models are considerably easier to solve than others.) Try to keep formulas short and simple. If a complicated formula is required, break it out into intermediate calculations with subtotals. For example, in the Everglade spreadsheet, each element of the loan payments is broken out explicitly: LT Interest, ST Interest, LT Payback, and ST Payback. Some of these columns could have been combined (e.g., into two columns with LT Payments and ST Payments, or even into one column for all Loan Payments). However, this makes the formulas more complicated and also makes the model harder to test and debug. As laid out, the individual formulas for the loan payments are so simple that their values can be predicted easily without even looking at the formula. This simplifies the testing and debugging of the model. Use Range Names Range names make formulas much easier to interpret. One way to refer to a block of related cells (or even a single cell) in a spreadsheet formula is to use its cell address (e.g., L11:L21 or C3). However, when reading the formula, this requires looking at that part of the spreadsheet to see what kind of information is given there. As mentioned previously in Sections 1.2 and 2.2, a better alternative is to assign a descriptive range name to the block of cells that immediately identifies what is there. (This is done by selecting the block of cells, clicking on the name box on the left of the formula bar above the spreadsheet, and then typing a name.) This is especially helpful when writing a formula for an output cell. Writing the formula in terms of range names instead of cell addresses makes the formula much easier to interpret. Range names also make the description of the model in Solver much easier to understand. Figure 4.5 illustrates the use of range names for the Everglade spreadsheet model. For example, consider the formula for long-term interest in cell F12. Since the long-term rate is given in cell C3 and the long-term loan amount is in cell D11, the formula for the longterm interest could have been written as = − C3*D11. However, by using the range name LTRate for cell C3 and the range name LTLoan for cell D11, the formula instead becomes = − LTRate*LTLoan, which is much easier to interpret at a glance. On the other hand, be aware that it is easy to get carried away with defining range names. Defining too many range names can be more trouble than it is worth. For example, when related data are grouped together in a table, we recommend giving a range name only for the entire table rather than for the individual rows and columns. In general, we suggest defining range names only for each group of data cells, the changing cells, the objective cell, and both sides of each group of constraints (the left-hand side and the right-hand side). Care also should be taken to assure that it is easy to quickly identify which cells are referred to by a particular range name. Use a name that corresponds exactly to the label on the spreadsheet. For example, in Figure 4.5, columns J and L are labeled Ending Balance 140 Chapter Four The Art of Modeling with Spreadsheets Spaces are not allowed in range names. When a range name has more than one word, we have used capital letters to distinguish the start of each new word in a range name (e.g., MinimumBalance). Another way is to use the underscore character (e.g., Minimum_Balance). and Minimum Balance on the spreadsheet, so we use the range names EndingBalance and Minimum­Balance. Using exactly the same name as the label on the spreadsheet makes it quick and easy to find the cells that are referred to by a range name. When desired, a list of all the range names and their corresponding cell addresses can be pasted directly into the spreadsheet by choosing Paste from the Use in Formula menu on the Formulas tab, and then clicking Paste List. Such a list (after reformatting) is included below essentially all the spreadsheets displayed in this text. When modifying an existing model that utilizes range names, care should be taken to assure that the range names continue to refer to the correct range of cells. When inserting a row or column into a spreadsheet model, it is helpful to insert the row or column into the middle of a range rather than at the end. For example, to add another product to a product-mix model with four products, add a column between Products 2 and 3 rather than after Product 4. This will automatically extend the relevant range names to span across all five columns since these range names will continue to refer to everything between Product 1 and Product 4, including the newly inserted column for the fifth product. Similarly, deleting a row or column from the middle of a range will contract the span of the relevant range names appropriately. You can double-check the cells that are referred to by a range name by choosing that range name from the name box (on the left of the formula bar above the spreadsheet). This will highlight the cells that are referred to by the chosen range name. Use Relative and Absolute References to Simplify Copying Formulas Excel’s fill commands provide a quick and reliable way to replicate a formula into multiple cells. Whenever multiple related formulas will be needed, try to enter the formula just once and then use Excel’s fill commands to replicate the formula. Not only is this quicker than retyping the formula, but it is also less prone to error. We saw a good example of this when discussing the expansion of the model to full-scale size in the preceding section. Starting with the two-year spreadsheet in Figure 4.3, fill commands were used to copy the formulas in columns F, G, I, J, and L for the remaining years to create the full-scale, 10-year spreadsheet in Figure 4.4. Using relative and absolute references for related formulas not only aids in building a model but also makes it easier to modify an existing model or template. For example, suppose that you have formulated a spreadsheet model for a product-mix problem but now wish to modify the model to add another resource and its constraint. This requires inserting a row into the spreadsheet. If the output cells are written with proper relative and absolute references, then it is simple to copy the existing formulas into the inserted row. Use Borders, Shading, and Colors to Distinguish between Cell Types Make it easy to spot all the cells of the same type. It is important to be able to easily distinguish between the data cells, changing cells, output cells, and objective cell in a spreadsheet. One way to do this is to use different borders and cell shading for each of these different types of cells. In the text, data cells appear lightly shaded, changing cells are shaded a medium amount with a light border, output cells appear with no shading, and the objective cell is shaded darkly with a heavy border. In the spreadsheet files for this chapter, data cells are light blue, changing cells are yellow, and the objective cell is orange. Obviously, you may use any scheme that you like. The important thing is to be consistent, so that you can quickly recognize the types of cells. Then, when you want to examine the cells of a certain type, the color will immediately guide you there. Show the Entire Model on the Spreadsheet Solver uses a combination of the spreadsheet and the Solver dialog box (or the model pane in Analytic Solver) to specify the model to be solved. Therefore, it is possible to include certain elements of the model (such as the ≤, =, or ≥ signs and/or the right-hand sides of the constraints) in Solver without displaying them in the spreadsheet. However, we strongly recommend that every element of the model be displayed on the spreadsheet. Every person using or adapting the model, or referring back to it later, needs to be able to interpret the model. This is much easier to do by viewing the model on the spreadsheet than by trying to decipher it from Solver. Furthermore, a printout of the spreadsheet does not include information from Solver. 4.3 Some Guidelines for Building “Good” Spreadsheet Models 141 Display every element of the model on the spreadsheet rather than relying on only Solver to include certain elements. In particular, all the elements of a constraint should be displayed on the spreadsheet. For each constraint, three adjacent cells should be used for the total of the left-hand side, the ≤, =, or ≥ sign in the middle, and the right-hand side. (Note in Figure 4.5 that this was done in columns J, K, and L of the spreadsheet for the Everglade problem.) As mentioned earlier, the changing cells and objective cell should be highlighted in some manner (e.g., with borders and/or cell shading). A good test is that you should not need to go to Solver to determine any element of the model. You should be able to identify the changing cells, the objective cell, and all the constraints in the model just by looking at the spreadsheet. A Poor Spreadsheet Model It is certainly possible to set up a linear programming spreadsheet model without utilizing any of these ideas. Figure 4.7 shows an alternative spreadsheet formulation for the Everglade problem that violates nearly every one of these guidelines. This formulation can still be solved using Solver, which in fact yields the same optimal solution as in Figure 4.5. However, the formulation has many problems. It is not clear which cells yield the solution (borders and/ or shading are not used to highlight the changing cells and objective cell). Without going to Solver, the constraints in the model cannot be identified (the spreadsheet does not show the entire model). The spreadsheet also does not show most of the data. For example, to determine the data used for the projected cash flows, the interest rates, or the starting balance, you FIGURE 4.7 A poor formulation of the spreadsheet model for the Everglade cash flow management problem. A 1 B C D E A Poor Formulation of the Everglade Cash Flow Problem 2 LT ST Ending 4 Year Loan Loan Balance 5 2018 4.65 2.85 0.50 6 2019 5.28 0.50 7 2020 9.88 0.50 8 2021 7.81 0.50 9 2022 2.59 0.50 10 2023 0 0.50 11 2024 4.23 0.50 12 2025 0 2.74 13 2026 0 0.51 14 2027 0 10.27 15 2028 3 Solver Parameters Set Objective Cell: E15 To: Max By Changing Variable Cells: C5, D5:D14 Subject to the Constraints: E5:E15 >= 0.5 Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 5.39 E 3 Ending 4 Balance 5 6 =1–8+C5+D5 7 =E6–4+D7–$C$5*(0.05)–D6*(1.07) 8 =E7+3+D8–$C$5*(0.05)–D7*(1.07) 9 =E8+6+D9–$C$5*(0.05)–D8*(1.07) 10 =E9+3+D10–$C$5*(0.05)–D9*(1.07) 11 =E10–4+D11–$C$5*(0.05)–D10*(1.07) 12 =E11+7+D12–$C$5*(0.05)–D11*(1.07) 13 =E12–2+D13–$C$5*(0.05)–D12*(1.07) 14 =E13+10+D14–$C$5*(0.05)–D13*(1.07) 15 =E14+D15–$C$5*(1.05)–D14*(1.07) =E5–2+D6–$C$5*(0.05)–D5*(1.07) F 142 Chapter Four The Art of Modeling with Spreadsheets need to dig into the formulas in column E (the data are not separate from the formulas). If any of these data change, the actual formulas need to be modified rather than simply changing a number on the spreadsheet. Furthermore, the formulas and the model in Solver are difficult to interpret (range names are not utilized). Compare Figures 4.5 and 4.7. Applying the guidelines for good spreadsheet models (as is done for Figure 4.5) results in a model that is easier to understand, easier to debug, and easier to modify. This is especially important for models that will have a long life span. If this model is going to be reused months later, the “good” model of Figure 4.5 immediately can be understood, modified, and reapplied as needed, whereas deciphering the spreadsheet model of Figure 4.7 again would be a great challenge. Review Questions 4.4 1. Which part of the model should be entered first on the spreadsheet? 2. Should numbers be included in formulas or entered separately in data cells? 3. How do range names make formulas and the model in Solver easier to interpret? How should range names be chosen? 4. What are some ways to distinguish data cells, changing cells, output cells, and objective cells on a spreadsheet? 5. How many cells are needed to completely specify a constraint on a spreadsheet? DEBUGGING A SPREADSHEET MODEL Debugging a spreadsheet model sometimes is as challenging as debugging a computer program. If you have added rows or columns to the spreadsheet, make sure that each of the range names still refers to the correct cells. Excel Tip: Pressing ­control-~ on a PC (or ­command-~ on a Mac) toggles the worksheet between viewing values and viewing formulas in all the output cells. Excel’s auditing tools enable you to either trace forward or backward to see the linkages between cells. No matter how carefully it is planned and built, even a moderately complicated model usually will not be error-free the first time it is run. Often the mistakes are immediately obvious and quickly corrected. However, sometimes an error is harder to root out. Following the guidelines in Section 4.3 for developing a good spreadsheet model can make the model much easier to debug. Even so, much like debugging a computer program, debugging a spreadsheet model can be a difficult task. This section presents some tips and a variety of Excel features that can make debugging easier. As a first step in debugging a spreadsheet model, test the model using the principles discussed in the first subsection on testing in Section 4.2. In particular, try different values for the changing cells for which you can predict the correct result in the output cells and see if they calculate as expected. Values of 0 are good ones to try initially because usually it is then obvious what should be in the output cells. Try other simple values, such as all 1s, where the correct results in the output cells are reasonably obvious. For more complicated values, break out a calculator and do some manual calculations to check the various output cells. Include some very large values for the changing cells to ensure that the calculations are behaving reasonably for these extreme cases. If you have defined range names, be sure that they still refer to the correct cells. Sometimes they can become disjointed when you add rows or columns to the spreadsheet. To test the range names, you can either select the various range names in the name box, which will highlight the selected range in the spreadsheet, or paste the entire list of range names and their references into the spreadsheet. Carefully study each formula to be sure it is entered correctly. A very useful feature in Excel for checking formulas is the toggle to switch back and forth between viewing the formulas in the worksheet and viewing the resulting values in the output cells. By default, Excel shows the values that are calculated by the various output cells in the model. Typing control-~ switches the current worksheet to instead display the formulas in the output cells, as shown in Figure 4.8. Typing control-~ again switches back to the standard view of displaying the values in the output cells (like Figure 4.5). Another useful set of features built into Excel are the auditing tools. The auditing tools are available in the Formula Auditing group of the Formulas Tab. The auditing tools can be used to graphically display which cells make direct links to a given cell. For example, selecting LTLoan (D11) in Figure 4.5 and then Trace Dependents generates the arrows on the spreadsheet shown in Figure 4.9. LT Rate ST Rate 4 –2 –4 3 3 –4 7 –2 10 Year 2018 2019 2020 2021 2022 2023 2024 2025 2026 2027 2028 10 D E 11 12 13 14 15 16 17 18 19 20 21 LT ST 6 G ST =–STRate*E15 =–LTRate*LTLoan 0 0 0 =–LTRate*LTLoan =–LTRate*LTLoan =–LTRate*LTLoan =–LTRate*LTLoan 4.23256 =–LTRate*LTLoan =–STRate*E20 =–STRate*E18 =–STRate*E19 =–STRate*E17 =–STRate*E16 =–STRate*E14 2.58639 =–LTRate*LTLoan 0 =–STRate*E13 7.80732 =–LTRate*LTLoan =–STRate*E12 9.88295 =–LTRate*LTLoan Interest =–STRate*E11 Interest LT (all cash figures in millions of dollars) F 5.28073 =–LTRate*LTLoan Flow Loan Loan –8 4.65124 2.84759 Cash 0.5 Start Balance Minimum Cash 7 8 9 1 0.07 0.05 6 5 C Everglade Cash Flow Management Problem B 2 3 1 A =–E18 =–E19 =–E17 =–E16 =–E15 =–E14 =–E13 =–E12 =–E11 Payback ST I =–LTLoan =–E20 Payback LT H =J20+SUM(C21:I21) =J18+SUM(C19:I19) =J19+SUM(C20:I20) =J17+SUM(C18:I18) =J16+SUM(C17:I17) =J15+SUM(C16:I16) =J14+SUM(C15:I15) =J13+SUM(C14:I14) =J12+SUM(C13:I13) =J11+SUM(C12:I12) Balance =StartBalance+SUM(C11:I11) Ending J ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ K =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash =MinimumCash Balance =MinimumCash Minimum L The spreadsheet obtained by toggling the spreadsheet in Figure 4.5 once to replace the values in the output cells by the formulas entered into those cells. Using the toggle feature in Excel once more will restore the view of the spreadsheet shown in Figure 4.5. FIGURE 4.8 4.4 Debugging a Spreadsheet Model 143 144 Chapter Four The Art of Modeling with Spreadsheets FIGURE 4.9 The spreadsheet obtained by using the Excel auditing tools to trace the dependents of the LT Loan value in cell D11 of the spreadsheet in Figure 4.5. A 1 B C D E F G H I J K L Everglade Cash Flow Management Problem 2 3 LT Rate 5% 4 ST Rate 7% (all cash figures in millions of dollars) 5 6 Start Balance 1 7 Minimum Cash 0.5 Cash LT ST LT ST LT ST Ending 10 Year Flow Loan Loan Interest Interest Payback Payback Balance 11 2018 –8 4.65 2.85 8 9 Minimum Balance 0.50 ≥ 0.5 –2 5.28 –0.23 –0.20 –2.85 0.50 ≥ 0.5 –4 9.88 –0.23 –0.37 –5.28 0.50 ≥ 0.5 2021 3 7.81 –0.23 –0.69 –9.88 0.50 ≥ 0.5 15 2022 6 2.59 –0.23 –0.55 –7.81 0.50 ≥ 0.5 16 2023 3 0 –0.23 –0.18 –2.59 0.50 ≥ 0.5 0.5 12 2019 13 2020 14 17 2024 –4 4.23 –0.23 0 0 0.50 ≥ 18 2025 7 0 –0.23 –0.30 –4.23 2.74 ≥ 0.5 19 2026 –2 0 0.5 20 2027 10 0 21 2028 The Evaluate Formula auditing tool can be used to evaluate a formula one step at a time. (At press time, this particular auditing tool is only available in Windows versions of Excel.) –0.23 –0.23 0 0 0.51 ≥ 0 0 10.27 ≥ 0.5 –0.23 0 0 5.39 ≥ 0.5 –4.65 You now can immediately see that LTLoan (D11) is used in the calculation of LT Interest for every year in column F, in the calculation of LTPayback (H21), and in the calculation of the ending balance in 2018 (J11). This can be very illuminating. Think about what output cells LTLoan should impact directly. There should be an arrow to each of these cells. If, for example, LTLoan is missing from any of the formulas in column F, the error will be immediately revealed by the missing arrow. Similarly, if LTLoan is mistakenly entered in any of the short-term loan output cells, this will show up as extra arrows. You also can trace backward to see which cells provide the data for any given cell. These can be displayed graphically by choosing Trace Precedents. For example, choosing Trace Precedents for the ST Interest cell for 2019 (G12) displays the arrows shown in Figure 4.10. These arrows indicate that the ST Interest cell for 2019 (G12) refers to the ST Loan in 2018 (E11) and to STRate (C4). When you are done, choose Remove Arrows. The auditing tools can also be used to see how each component of a formula evaluates, one step at a time. For example, selecting the objective cell EndBalance (J21) and then Evaluate Formula brings up the dialog box shown in Figure 4.11. Inside the evaluation box is the formula in the selected cell, =J20 + SUM(C21:I21), with one component of the formula underlined. Pressing the Evaluate button evaluates the underlined component of the formula (J20 becomes 10.273...) and then underlines the next component of the formula, SUM(C21:I21). As shown at the bottom of the figure, each press of the Evaluate button evaluates the underlined component of the formula and then underlines the next component of the formula to be evaluated. Three presses fully evaluates the formula to show the end result of 5.39. This tool can be particularly useful for evaluating complicated formulas to make sure they are set up correctly. 4.4 Debugging a Spreadsheet Model 145 FIGURE 4.10 The spreadsheet obtained by using the Excel auditing tools to trace the precedents of the ST Interest (2019) calculation in cell G12 of the spreadsheet in Figure 4.5. A 1 B C D E F G H I J K L Everglade Cash Flow Management Problem 2 3 LT Rate 5% 4 ST Rate 7% (all cash figures in millions of dollars) 5 6 Start Balance 1 7 Minimum Cash 0.5 Cash LT ST LT ST LT ST Ending 10 Year Flow Loan Loan Interest Interest Payback Payback Balance 11 2018 –8 4.65 2.85 8 9 –0.23 ≥ 0.5 –0.23 –0.37 –5.28 0.50 ≥ 0.5 –0.69 –9.88 0.50 ≥ 0.5 –0.23 –0.55 –7.81 0.50 ≥ 0.5 –0.23 –0.18 –2.59 0.50 ≥ 0.5 0 0 0.50 ≥ 0.5 –0.30 –4.23 2.74 ≥ 0.5 0.5 –2 5.28 2020 –4 9.88 14 2021 3 7.81 –0.23 15 2022 6 2.59 16 2023 3 0 17 2024 –4 4.23 –0.23 18 2025 7 0 –0.23 –2 0 2027 10 0 21 2028 0.5 0.50 2019 2026 ≥ –2.85 13 20 Balance 0.50 –0.20 12 19 Minimum –0.23 –0.23 0 0 0.51 ≥ 0 0 10.27 ≥ 0.5 –0.23 0 0 5.39 ≥ 0.5 –4.65 FIGURE 4.11 The dialog box for the Evaluate Formula auditing tool when applied to the formula in EndBalance (J21). Successive presses of the Evaluate button evaluate the underlined portion of the formula, step-by-step, as shown below the dialog box. Review Questions 1. What is a good first step for debugging a spreadsheet model? 2. How do you toggle between viewing formulas and viewing values in output cells? 3. Which Excel tools can be used to trace the dependents or precedents for a given cell or evaluate a formula in a cell? 146 Chapter Four The Art of Modeling with Spreadsheets 4.5 Summary There is a considerable art to modeling well with spreadsheets. This chapter focuses on providing a foundation for learning this art. The general process of modeling in spreadsheets has four major steps: (1) plan the spreadsheet model, (2) build the model, (3) test the model, and (4) analyze the model and its results. During the planning step, it is helpful to begin by visualizing where you want to finish and then doing some calculations by hand to clarify the needed computations before starting to sketch out a logical layout for the spreadsheet. Then, when you are ready to undertake the building step, it is a good idea to start by building a small, readily manageable version of the model before expanding the model to full-scale size. This enables you to test the small version first to get all the logic straightened out correctly before expanding to a fullscale model and undertaking a final test. After completing all of this, you are ready for the analysis step, which involves applying the model to evaluate proposed solutions and perhaps using Solver to optimize the model. Using this plan-build-test-analyze process should yield a spreadsheet model, but it doesn’t guarantee that you will obtain a good one. Section 4.3 describes in detail the following guidelines for building “good” spreadsheet models: • • • • • • • • • Enter the data first. Organize and clearly identify the data. Enter each piece of data into one cell only. Separate data from formulas. Keep it simple. Use range names. Use relative and absolute references to simplify copying formulas. Use borders, shading, and colors to distinguish between cell types. Show the entire model on the spreadsheet. Even if all these guidelines are followed, a thorough debugging process may be needed to eliminate the errors that lurk within the initial version of the model. It is important to check whether the output cells are giving correct results for various values of the changing cells. Other items to check include whether range names refer to the appropriate cells and whether formulas have been entered into output cells correctly. Excel provides a number of useful features to aid in the debugging process. One is the ability to toggle the worksheet between viewing the results in the output cells and the formulas entered into those output cells. Several other helpful features are available with Excel’s auditing tools. Glossary auditing tools A set of tools provided by Excel to aid in debugging a spreadsheet model. (Section 4.4), 142 range name A descriptive name given to a range of cells that immediately identifies what is there. (Section 4.3), 139 toggle The act of switching back and forth between viewing the results in the output cells and viewing the formulas entered into those ­output cells. (Section 4.4), 142 Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Excel Files: Excel Add-in: Everglade Case Study Analytic Solver Wyndor Example Everglade Problem 4.12 Everglade Problem 4.13 Solved Problems The solutions are available at www.mhhe.com/Hillier6e. 4.S1. Production and Inventory Planning Model Surfs Up produces high-end surfboards. A challenge faced by Surfs Up is that their demand is highly seasonal. Demand exceeds production capacity during the warm summer months, but is very low in the winter months. To meet the high demand during the summer, Surfs Up typically produces more surfboards Chapter 4 Problems 147 than are needed in the winter months and then carries inventory into the summer months. Their production facility can produce at most 50 boards per month using regular labor at a cost of $125 each. Up to 10 additional boards can be produced by utilizing overtime labor at a cost of $135 each. The boards are sold for $200. Because of storage cost and the opportunity cost of capital, each board held in inventory from one month to the next incurs a cost of $5 per board. Since demand is uncertain, Surfs Up would like to maintain an ending inventory (safety stock) of at least 10 boards during the warm months (May–September) and at least 5 boards during the other months (October–April). It is now the start of January and Surfs Up has 5 boards in inventory. The forecast of demand over the next 12 months is shown in the following table. Formulate and solve a linear programming model in a spreadsheet to determine how many surfboards should be produced each month to maximize total profit. Forecasted Demand Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. 10 14 15 20 45 65 85 85 40 30 15 15 4.S2. Aggregate Planning: Manpower Hiring/ Firing/Training Cool Power produces air-conditioning units for large commercial properties. Because of the low cost and efficiency of its products, the company has been growing from year to year. Also, seasonality in construction and weather conditions create production requirements that vary from month to month. Cool Power currently has 10 fully trained employees working in manufacturing. Each trained employee can work 160 hours per month and is paid a monthly wage of $4,000. New trainees can be hired at the beginning of any month. Because of their lack of initial skills and required training, a new trainee provides only 100 hours of useful labor in the first month, but is still paid a full monthly wage of $4,000. Furthermore, because of required interviewing and training, there is a $2,500 hiring cost for each employee hired. After one month, a trainee is considered fully trained. An employee can be fired at the beginning of any month, but must be paid two weeks of severance pay ($2,000). Over the next 12 months, Cool Power forecasts the labor requirements shown in the following table. Since management anticipates higher requirements next year, Cool Power would like to end the year with at least 12 fully trained employees. How many trainees should be hired and/or workers fired in each month to meet the labor requirements at the minimum possible cost? Formulate and solve a linear programming spreadsheet model. Labor Requirements (hours) Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. 1,600 2,000 2,000 2,000 2,800 3,200 3,600 3,200 1,600 1,200 800 800 Problems An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 4.1. Consider the Everglade cash flow problem discussed in this chapter. Suppose that extra cash is kept in an interest-­ bearing savings account. Assume that any cash left at the end of a year earns 3 percent interest the following year. Make any necessary modifications to the spreadsheet and re-solve. (The original spreadsheet for this problem is available among the Excel files for this chapter in www.mhhe.com/Hillier6e.) 4.2.* The Pine Furniture Company makes fine country furniture. The company’s current product lines consist of end tables, coffee tables, and dining room tables. The production of each of these tables requires 8, 15, and 80 pounds of pine wood, respectively. The tables are handmade and require one hour, two hours, and four hours, respectively. Each table sold generates $50, $100, and $220 profit, respectively. The company has 3,000 pounds of pine wood and 200 hours of labor available for the coming week’s production. The chief operating officer (COO) has asked you to do some spreadsheet modeling with these data to analyze what the product mix should be for the coming week and make a recommendation. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will the COO need? b. Suppose that Pine Furniture were to produce three end tables and three dining room tables. Calculate by hand the amount of pine wood and labor that would be required, as well as the profit generated from sales. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model and then solve it. 4.3. Reboot, Inc., is a manufacturer of hiking boots. Demand for boots is highly seasonal. In particular, the demand in the next year is expected to be 3,000, 4,000, 8,000, and 7,000 pairs of boots in quarters 1, 2, 3, and 4, respectively. With its current production facility, the company can produce at most 6,000 pairs of boots in any quarter. Reboot would like to meet all the expected demand, so it will need to carry inventory to meet demand in the later quarters. Each pair of boots sold generates a profit of $20 per pair. Each pair of boots in inventory at the end of a quarter incurs $8 in storage and capital recovery costs. Reboot has 1,000 pairs of boots in inventory at the start of quarter 1. Reboot’s top management has given you the assignment of doing some spreadsheet modeling to analyze what the production schedule should be for the next four quarters and making a recommendation. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers are needed? 148 Chapter Four The Art of Modeling with Spreadsheets b. Suppose that Reboot were to produce 5,000 pairs of boots in each of the first two quarters. Calculate by hand what the ending cash positions would be after year 1 and year 2. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for years 1 and 2, and then thoroughly test the model. e. Expand the model to full scale and then solve it. 4.4.* The Fairwinds Development Corporation is considering taking part in one or more of three different development ­projects—A, B, and C—that are about to be launched. Each project requires a significant investment over the next few years and then would be sold upon completion. The projected cash flows (in millions of dollars) associated with each project are shown in the table below. Year 1 2 3 4 5 6 Project A Project B Project C −4 −6 −6 24 0 0 −8 −8 −4 −4 30 0 −10 −7 −7 −5 23 44 Fairwinds has $10 million available now and expects to receive $6 million from other projects by the end of each year (1 through 6) that would be available for the ongoing investments the following year in projects A, B, and C. By acting now, the company may participate in each project either fully, fractionally (with other development partners), or not at all. If Fairwinds participates at less than 100 percent, then all the cash flows associated with that project are reduced proportionally. Company policy requires ending each year with a cash balance of at least $1 million. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will top management need? b. Suppose that Fairwinds were to participate in Project A fully and in Project C at 50 percent. Calculate by hand the ending inventory, profit from sales, and inventory costs for quarters 1 and 2. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for quarters 1 and 2, and then thoroughly test the model. e. Expand the model to full scale, and then solve it. 4.5. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 4.3. Briefly describe how spreadsheet modeling was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 4.6. Decorum, Inc., manufactures high-end ceiling fans. Their sales are seasonal with higher demand in the warmer summer months. Typically, sales average 400 units per month. However, in the hot summer months (June, July, and August), sales spike up to 600 units per month. Decorum can produce up to 500 units per month at a cost of $300 each. By bringing in temporary workers, Decorum can produce up to an additional 75 units at a cost of $350 each. Decorum sells the ceiling fans for $500 each. Decorum can carry inventory from one month to the next, but at a cost of $20 per ceiling fan per month. Decorum has 25 units in inventory at the start of January. Assuming Decorum must produce enough ceiling fans to meet demand, how many ceiling fans should Decorum produce each month (using their regular labor force and/or temporary workers) over the course of the next year so as to maximize their total profit? a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will Decorum require? b. Suppose Decorum builds 450 ceiling fans in January and 550 (utilizing temporary workers) in February. Calculate by hand the total costs for January and February. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for January and February and thoroughly test the model. e. Build and solve a linear programming spreadsheet model to maximize the profit over all 12 months. 4.7. Allen Furniture is a manufacturer of hand-crafted furniture. At the start of January, Allen employs 20 trained craftspeople. They have forecasted their labor needs over the next 12 months as shown in the following table. Each trained craftsperson provides 200 labor-hours per month and is paid a wage of $3,000 per month. Hiring new craftspeople requires advertising, interviewing, and then training at a cost of $2,500 per hire. New hires are called apprentices for their first month. Apprentices spend their first month observing and learning. They are paid $2,000 for the month, but provide no labor. In their second month, apprentices are reclassified as trained craftspeople. The union contract allows for firing a craftsperson at the beginning of a month, but $1,500 must be given in severance pay. Moreover, at most 10% of the trained craftspeople can be fired in any month. Allen would like to start next year with at least 25 trained craftspeople. How many apprentices should be hired and how many craftspeople should be fired in each month to meet the labor requirements at the minimum possible cost? a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will Allen require? b. Suppose one apprentice is hired in January. Calculate by hand how many labor-hours would be available in January and February. Calculate by hand the total costs in January and February. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. Labor Requirements (hours) Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. 3,400 4,000 4,200 4,200 3,000 2,800 3,000 4,000 4,500 5,000 5,200 4,800 Chapter 4 Problems 149 d. Build a spreadsheet model for January and February and thoroughly test the model. e. Build and solve a linear programming spreadsheet model to maximize the profit over all 12 months. 4.8. Refer to the scenario described in Problem 3.13 (­ Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Web Mercantile’s problem by doing the following. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will Web Mercantile require? b. Suppose that Web Mercantile were to lease 30,000 square feet for all five months and then 20,000 additional square feet for the last three months. Calculate the total costs by hand. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for months 1 and 2, and then thoroughly test the model. e. Expand the model to full scale, and then solve it. 4.9.* Refer to the scenario described in Problem 3.15 (­ Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Larry Edison’s problem by doing the following. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What are the numbers that Larry will require? b. Suppose that Larry were to hire three full-time workers for the morning shift, two for the afternoon shift, and four for the evening shift, as well as three A 1 part-time workers for each of the four shifts. Calculate by hand how many workers would be working at each time of the day and what the total cost would be for the entire day. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model and then solve it. 4.10. Refer to the scenario described in Problem 3.19 (­Chapter 3), but ignore the instructions given there. Focus instead on using spreadsheet modeling to address Al Ferris’s problem by doing the following. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers will Al require? b. Suppose that Al were to invest $20,000 each in investment A (year 1), investment B (year 2), and investment C (year 2). Calculate by hand what the ending cash position would be after each year. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for years 1 through 3, and then thoroughly test the model. e. Expand the model to full scale, and then solve it. 4.11. In contrast to the spreadsheet model for the Wyndor Glass Co. product-mix problem shown in Figure 4.6, the following spreadsheet is an example of a poorly formulated spreadsheet model for this same problem. Referring to Section 4.3, identify the guidelines violated by the model below. Then, explain how each guideline has been violated and why the model in Figure 4.6 is a better alternative. B C D Wyndor Glass Co. (Poor Formulation) 2 3 Doors Produced 2 4 Windows Produced 6 5 Hours Used (Plant 1) 2 6 Hours Used (Plant 2) 12 7 Hours Used (Plant 3) 18 8 Total Profit Solver Parameters Set Objective Cell: C8 To: Max By Changing Variable Cells: C3:C4 Subject to the Constraints: C5 <= 4 C6 <= 12 C7 <= 18 Solver Options: Make Variables Nonnegative Solving Method: Simplex LP $3,600 B C 5 Hours Used (Plant 1) =1*C3+0*C4 6 Hours Used (Plant 2) =0*C3+2*C4 7 8 Hours Used (Plant 3) =3*C3+2*C4 Total Profit =300*C3+500*C4 150 Chapter Four The Art of Modeling with Spreadsheets 4.12. Refer to the spreadsheet file named “Everglade Problem 4.12,” available in www.mhhe.com/Hillier6e. This file contains a formulation of the Everglade problem considered in this chapter. However, three errors are included in this formulation. Use the ideas presented in Section 4.4 for debugging a spreadsheet model to find the errors. In particular, try different trial values for which you can predict the correct results, use the toggle to examine all the formulas, and use the auditing tools to check precedence and dependence relationships among the various changing cells, data cells, and output cells. Describe the errors found and how you found them. 4.13. Refer to the spreadsheet file named “Everglade Problem 4.13,” available in www.mhhe.com/Hillier6e. This file contains a formulation of the Everglade problem considered in this chapter. However, three errors are included in this formulation. Use the ideas presented in Section 4.4 for debugging a spreadsheet model to find the errors. In particular, try different trial values for which you can predict the correct results, use the toggle to examine all the formulas, and use the auditing tools to check precedence and dependence relationships among the various changing cells, data cells, and output cells. Describe the errors found and how you found them. Case 4-1 Prudent Provisions for Pensions Among its many financial products, the Prudent Financial Services Corporation (normally referred to as PFS) manages a wellregarded pension fund that is used by a number of companies to provide pensions for their employees. PFS’s management takes pride in the rigorous professional standards used in operating the fund. Since the near-collapse of the financial markets during the protracted Great Recession that began in late 2007, PFS has redoubled its efforts to provide prudent management of the fund. It is now December 2017. The total pension payments that will need to be made by the fund over the next 10 years are shown in the following table. Year Pension Payments ($ millions) 2018 2019 2020 2021 2022 2023 2024 2025 2026 2027 8 12 13 14 16 17 20 21 22 24 By using interest as well, PFS currently has enough liquid assets to meet all these pension payments. Therefore, to safeguard the pension fund, PFS would like to make a number of investments whose payouts would match the pension payments over the next 10 years. The only investments that PFS trusts for the pension fund are a money market fund and bonds. The money market fund pays an annual interest rate of 2 percent. The characteristics of each unit of the four bonds under consideration are shown in the table below. Bond 1 Bond 2 Bond 3 Bond 4 Current Price Coupon Rate Maturity Date Face Value $980 920 750 800 4% 2 0 3 Jan. 1, 2019 Jan. 1, 2021 Jan. 1, 2023 Jan. 1, 2026 $1,000 1,000 1,000 1,000 All of these bonds will be available for purchase on January 1, 2018, in as many units as desired. The coupon rate is the percentage of the face value that will be paid in interest on January 1 of each year, starting one year after purchase and continuing until (and including) the maturity date. Thus, these interest payments on January 1 of each year are in time to be used toward the pension payments for that year. Any excess interest payments will be deposited into the money market fund. To be conservative in its financial planning, PFS assumes that all the pension payments for the year occur at the beginning of the year immediately after these interest payments (including a year’s interest from the money market fund) are received. The entire face value of a bond also will be received on its maturity date. Since the current price of each bond is less than its face value, the actual yield of the bond exceeds its coupon rate. Bond 3 is a zero-­coupon bond, so it pays no interest but instead pays a face value on the maturity date that greatly exceeds the purchase price. PFS would like to make the smallest possible investment (including any deposit into the money market fund) on January 1, 2018, to cover all its required pension payments through 2027. Some spreadsheet modeling needs to be done to see how to do this. a. Visualize where you want to finish. What are the decisions that need to be made? What should the objective be? What numbers are needed by PFS management? b. Suppose that PFS were to invest $30 million in the money market fund and purchase 10,000 units each of bond 1 and bond 2 on January 1, 2018. Calculate by hand the payments received from bonds 1 and 2 on January 1 of 2019 and 2020. Also calculate the resulting balance in the money market fund on January 1 of 2018, 2019, and 2020 after receiving these payments, making the pension payments for the year, and depositing any excess into the money market fund. c. Make a rough sketch of a spreadsheet model, with blocks laid out for the data cells, changing cells, output cells, and objective cell. d. Build a spreadsheet model for years 2018 through 2020, and then thoroughly test the model. e. Expand the model to consider all years through 2027, and then solve it. Additional Case Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. Chapter Five What-If Analysis for Linear Programming Learning Objectives After completing this chapter, you should be able to 1. Explain what is meant by what-if analysis. 2. Summarize the benefits of what-if analysis. 3. Enumerate the different kinds of changes in the model that can be considered by what-if analysis. 4. Describe how the spreadsheet formulation of the problem can be used to perform any of these kinds of what-if analysis. 5. Use parameter analysis reports generated by Analytic Solver to systematically investigate the effect of changing either one or two data cells to various other trial values. 6. Find how much any single coefficient in the objective function can change without changing the optimal solution. 7. Evaluate simultaneous changes in objective function coefficients to determine whether the changes are small enough that the original optimal solution must still be optimal. 8. Predict how the value in the objective cell would change if a small change were to be made in the right-hand side of one or more of the functional constraints. 9. Find how much the right-hand side of a single functional constraint can change before this prediction becomes no longer valid. 10. Evaluate simultaneous changes in right-hand sides to determine whether the changes are small enough that this prediction must still be valid. 11. Describe the goal of robust optimization and how it is implemented with independent parameters. 12. Use chance constraints to deal with constraints that actually can be violated a little bit. Chapters 2 to 4 have described and illustrated how to formulate a linear programming model on a spreadsheet to represent a variety of managerial problems, and then how to use Solver to find an optimal solution for this model. You might think that this would finish our story about linear programming: Once the manager learns the optimal solution, she would immediately implement this solution and then turn her attention to other matters. However, this is not the case. The enlightened manager demands much more from linear programming, and linear programming has much more to offer her—as you will discover in this chapter. An optimal solution is only optimal with respect to a particular mathematical model that provides only a rough representation of the real problem. A manager is interested in much more than just finding such a solution. The purpose of a linear programming study is to help guide management’s final decision by providing insights into the likely consequences of pursuing various managerial options under a variety of assumptions about future conditions. Most of the important insights are gained while conducting analysis after finding an optimal solution for the original version of the basic model. This analysis is commonly referred to as what-if analysis because it involves addressing some questions about what would happen to 151 152 Chapter Five What-If Analysis for Linear Programming the optimal solution if different assumptions were made about future conditions. Spreadsheets play a central role in addressing these what-if questions. This chapter focuses on the types of information provided by what-if analysis and why it is valuable to managers. The first section provides an overview. Section 5.2 returns to the Wyndor Glass Co. product-mix case study (Section 2.1) to describe the what-if analysis that is needed in this situation. The next four sections then perform this what-if analysis, using a variety of procedures that are applicable to any linear programming problem. Section 5.7 extends this approach further to provide a way of obtaining a solution that is virtually guaranteed to be feasible. Section 5.8 describes how to deal with constraints that actually can be violated a little bit without serious complications. 5.1 THE IMPORTANCE OF WHAT-IF ANALYSIS TO MANAGERS In real applications, many of the numbers in the model may be only rough estimates. What happens to the optimal solution if an error is made in estimating a parameter of the model? The examples and problems in the preceding chapters on linear programming have provided the data needed to determine precisely all the numbers that should go into the data cells for the spreadsheet formulation of the linear programming model. (Recall that these numbers are referred to as the parameters of the model.) Real applications seldom are this straightforward. Substantial time and effort often are needed to track down the needed data. Even then, it may be possible to develop only rough estimates of the parameters of the model. For example, in the Wyndor case study, two key parameters of the model are the coefficients in the objective function that represent the unit profits of the two new products. These parameters were estimated to be $300 for the doors and $500 for the windows. However, these unit profits depend on many factors—the costs of raw materials, production, shipping, advertising, and so on, as well as such things as the market reception to the new products and the amount of competition encountered. Some of these factors cannot be estimated with real accuracy until long after the linear programming study has been completed and the new products have been on the market for some time. Therefore, before Wyndor’s management makes a decision on the product mix, it will want to know what the effect would be if the unit profits turn out to differ significantly from the estimates. For example, would the optimal solution change if the unit profit for the doors turned out to be $200 instead of the estimate of $300? How inaccurate can the estimate be in either direction before the optimal solution changes? Such questions are addressed in Section 5.3 when only one estimate is inaccurate. ­Section 5.4 will address similar questions when multiple estimates are inaccurate. If the optimal solution will remain the same over a wide range of values for a particular coefficient in the objective function, then management will be content with a fairly rough estimate for this coefficient. On the other hand, if even a small error in the estimate would change the optimal solution, then management will want to take special care to refine this estimate. Management sometimes will get involved directly in adjusting such estimates to its satisfaction. Here then is a summary of the first benefit of what-if analysis: 1. Typically, many of the parameters of a linear programming model are only estimates of quantities (e.g., unit profits) that cannot be determined precisely at this time. What-if analysis reveals how close each of these estimates needs to be to avoid obtaining an erroneous optimal solution, and therefore pinpoints the sensitive parameters (those parameters where extra care is needed to refine their estimates because even small changes in their values can change the optimal solution). Several sections describe how what-if analysis provides this benefit for the most important parameters. Sections 5.3 and 5.4 do this for the coefficients in the objective function (these numbers typically appear in the spreadsheet in the row for the unit contribution of each activity toward the overall measure of performance). Sections 5.5 and 5.6 do the same for the right-hand sides of the functional constraints (these are the numbers that typically are in the right-hand column of the spreadsheet just to the right of the ≤, ≥, or = signs). Businesses operate in a dynamic environment. Even when management is satisfied with the current estimates and implements the corresponding optimal solution, conditions may change 5.1 The Importance of What-If Analysis to Managers 153 What happens to the optimal solution if conditions change in the future? What happens if managerial policy decisions change? later. For example, suppose that Wyndor’s management is satisfied with $300 as the estimate of the unit profit for the doors, but increased competition later forces a price reduction that reduces this unit profit. Does this change the optimal product mix? The what-if analysis shown in Section 5.3 immediately indicates in advance which new unit profits would leave the optimal product mix unchanged, which can help guide management in its new pricing decision. Furthermore, if the optimal product mix is unchanged, then there is no need to solve the model again with the new coefficient. Avoiding solving the model again is no big deal for the tiny twovariable Wyndor problem, but it is extremely welcome for real applications that may have hundreds or thousands of constraints and variables. In fact, for such large models, it may not even be practical to re-solve the model repeatedly to consider the many possible changes of interest. Thus, here is the second benefit of what-if analysis: 2. If conditions change after the study has been completed (a common occurrence), what-if analysis leaves signposts that indicate (without solving the model again) whether a resulting change in a parameter of the model changes the optimal solution. Again, several subsequent sections describe how what-if analysis does this. These sections focus on studying how changes in the parameters of a linear programming model affect the optimal solution. This type of what-if analysis commonly is referred to as sensitivity analysis, because it involves checking how sensitive the optimal solution is to the value of each parameter. Sensitivity analysis is a vital part of what-if analysis. However, rather than being content with the passive sensitivity analysis approach of checking the effect of parameter estimates being inaccurate, what-if analysis often goes further to take a proactive approach. An analysis may be made of various possible managerial actions that would result in changes to the model. A prime example of this proactive approach arises when certain parameters of the model represent managerial policy decisions rather than quantities that are largely outside the control of management. For example, for the Wyndor product-mix problem, the right-hand sides of the three functional constraints (4, 12, 18) represent the number of hours of production time in the three respective plants being made available per week for the production of the two new products. Management can change these three resource amounts by altering the production levels for the old products in these plants. Therefore, after learning the optimal solution, management will want to know the impact on the profit from the new products if these resource amounts are changed in certain ways. One key question is how much this profit can be increased by increasing the available production time for the new products in just one of the plants. Another is how much this profit can be increased by simultaneously making helpful changes in the available production times in all the plants. If the profit from the new products can be increased enough to more than compensate for the profit lost by decreasing the production levels for certain old products, management probably will want to make the change. We now can summarize the third benefit of what-if analysis: 3. When certain parameters of the model represent managerial policy decisions, what-if analysis provides valuable guidance to management regarding the impact of altering these policy decisions. Sections 5.5 and 5.6 will explore this benefit further. What-if analysis sometimes goes even further in providing helpful guidance to m ­ anagement. For example, when there is considerable uncertainty about what the actual values of the model parameters will turn out to be, management might want to identify a solution that is virtually guaranteed to be both feasible and nearly optimal for all plausible combinations of the actual values of these parameters. Section 5.7 will introduce a robust optimization technique for doing this. Section 5.8 then will describe how to use chance constraints to deal with constraints that can be violated a little bit without serious complications. Throughout this chapter, the focus is on how and when to use what-if analysis while applying linear programming. What-if analysis also is important to managers when applying the various other techniques of management science. Although the procedures may need to be adapted to fit other kinds of applications, this chapter illustrates the importance of including what-if analysis in almost any management science study. 154 Chapter Five What-If Analysis for Linear Programming Review Questions 1. 2. 3. 4. 5. 6. 7. 8. 5.2 What are the parameters of a linear programming model? How can inaccuracies arise in the parameters of a model? What does what-if analysis reveal about the parameters of a model that are only estimates? Is it always inappropriate to make only a fairly rough estimate for a parameter of a model? Why? How is it possible for the parameters of a model to be accurate initially and then become inaccurate at a later date? How does what-if analysis help management prepare for changing conditions? What is meant by sensitivity analysis? For what kinds of managerial policy decisions does what-if analysis provide guidance? CONTINUING THE WYNDOR CASE STUDY We now return to the case study introduced in Section 2.1 involving the Wyndor Glass Co. product-mix problem. To review briefly, recall that the company is preparing to introduce two new products: • An 8-foot glass door with aluminum framing. • A 4-foot × 6-foot double-hung wood-framed window. To analyze which mix of the two products would be most profitable, the company’s Management Science Group introduced two decision variables: D = Production rate of this new kind of door W = Production rate of this new kind of window where this rate measures the number of units produced per week. Three plants will be involved in the production of these products. Based on managerial decisions regarding how much these plants will continue to be used to produce current products, the number of hours of production time per week being made available in Plants 1, 2, and 3 for the new products is 4, 12, and 18, respectively. After obtaining rough estimates that the profit per unit will be $300 for the doors and $500 for the windows, the Management Science Group then formulated the linear programming model shown in Figure 2.12 and repeated here in Figure 5.1, where the objective is to choose the values of D and W in the changing cells UnitsProduced (C12:D12) so as to maximize the total profit (per week) given in the objective cell TotalProfit (G12). Applying Solver to this model yielded the optimal solution shown on this spreadsheet and summarized as follows. Optimal Solution D =2 ( Produce 2 doors per week.) W = 6 ( Produce 6 windows per week. ) Profit = 3,600 ( The estimated total weekly profit is $3,600.) However, this optimal solution assumes that all the estimates that provide the parameters of the model (as shown in the UnitProfit (C4:D4), HoursUsedPerUnitProduced (C7:D9), and HoursAvailable (G7:G9) data cells) are accurate. The head of the Management Science Group, Lisa Taylor, now is ready to meet with management to discuss the group’s recommendation that the above product mix be used. Management’s Discussion of the Recommended Product Mix Lisa Taylor (head of Management Science Group): I asked for this meeting so we could explore what questions the two of you would like us to pursue further. In particular, I am especially concerned that we weren’t able to better pin down just what the numbers should be to go into our model. Which estimates do you think are the shakiest? Bill Tasto (vice president for manufacturing): Without question, the estimates of the unit profits for the two products. Since the products haven’t gone into production yet, all we could do is analyze the data from similar current products and then try to project what 5.2 Continuing the Wyndor Case Study 155 FIGURE 5.1 The spreadsheet model and its optimal solution for the original Wyndor problem before beginning what-if analysis. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 2 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 2 6 $3,600 10 11 12 Units Produced Solver Parameters Set Objective Cell: TotalProfit To: Max By Changing Variable Cells: UnitsProduced Subject to the Constraints: HoursUsed <= HoursAvailable Solver Options: Make Variables Nonnegative Solving Method: Simplex LP E 5 6 7 8 9 Used =SUMPRODUCT(C7:D7, UnitsProduced) =SUMPRODUCT(C8:D8, UnitsProduced) =SUMPRODUCT(C9:D9, UnitsProduced) G 11 12 The allowable range for a unit profit indicates how far its estimate can be off without affecting the optimal product mix. Hours Total Profit =SUMPRODUCT(UnitProfit, UnitsProduced) Range Name Cells DoorsProduced HoursAvailable HoursUsed HoursUsedPerUnitProduced TotalProfit UnitProfit UnitsProduced WindowsProduced C12 G7:G9 E7:E9 C7:D9 G12 C4:D4 C12:D12 D12 the changes would be for these new products. We have some numbers, but they are pretty rough. We would need to do a lot more work to pin down the numbers better. John Hill (president): We may need to do that. Lisa, do you have a way of checking how far off one of these estimates can be without changing the optimal product mix? Lisa: Yes, we do. We can quickly find what we call the allowable range for each unit profit. As long as the true value of the unit profit is within this allowable range, and the other unit profit is correct, the optimal product mix will not change. If this range is pretty wide, you don’t need to worry about refining the estimate of the unit profit. However, if the range is quite narrow, then it is important to pin down the estimate more closely. John: What happens if both estimates are off? Lisa: We can provide a way of checking whether the optimal product mix might change for any new combination of unit profits you think might be the true one. John: Great. That’s what we need. There’s also another thing. Bill gave you the numbers for how many hours of production time we’re making available per week in the three plants for these new products. I noticed you used these numbers on your spreadsheet. Lisa: Yes. They’re the right-hand sides of our constraints. Is something wrong with these numbers? 156 Chapter Five What-If Analysis for Linear Programming John: No, but we would like your group to provide us with some analysis of what the effect would be if we change any one of those numbers. How much more profit could we get from the new products for each additional hour of production time per week we provide in one of the plants? That sort of thing. Lisa: Yes, we can get that analysis to you right away also. John: We might also be interested in changing the available production hours for two or three of the plants. Other key numbers that can only be estimated are the production rates of the doors and windows in plant 3. What happens if those estimates are uncertain? Lisa: No problem. We’ll give you information about all of that as well. Summary of Management’s What-If Questions Here is a summary of John Hill’s what-if questions that Lisa and her group will be addressing in the coming sections. 1. What happens if the estimate of the unit profit of one of Wyndor’s new products is inaccurate? (Section 5.3) 2. What happens if the estimates of the unit profits of both of Wyndor’s new products are inaccurate? (Section 5.4) 3. What happens if a change is made in the number of hours of production time per week being made available to Wyndor’s new products in one of the plants? (Section 5.5) 4. What happens if simultaneous changes are made in the number of hours of production time per week being made available to Wyndor’s new products in all the plants? (Section 5.6) 5. What happens if the production rates of doors and windows at plant 3 are uncertain? (­Sections 5.7 and 5.8) Review Questions 1. Which estimates of the parameters in the linear programming model for the Wyndor problem are most questionable? 2. Which numbers in this model represent tentative managerial decisions that management might want to change after receiving the Management Science Group’s analysis? 5.3 THE EFFECT OF CHANGES IN ONE OBJECTIVE FUNCTION COEFFICIENT Section 5.1 began by discussing the fact that many of the parameters of a linear programming model typically are only estimates of quantities that cannot be determined precisely at the time. What-if analysis (or sensitivity analysis in particular) reveals how close each of these estimates needs to be to avoid obtaining an erroneous optimal solution. We focus in this section on how sensitivity analysis does this when the parameters involved are coefficients in the objective function. (Recall that each of these coefficients gives the unit contribution of one of the activities toward the overall measure of performance.) In the process, we will address the first of the what-if questions posed by Wyndor management in the preceding section. Question 1: What happens if the estimate of the unit profit of one of Wyndor’s new products is inaccurate? To start this process, first consider the question of what happens if the estimate of $300 for the unit profit for Wyndor’s new kind of door is inaccurate. To address this question, let D = Unit profit for the new kind of door P = Cell C4 in the spreadsheet (see Figure 5.1) Although PD = $300 in the current version of Wyndor’s linear programming model, we now want to explore how much larger or how much smaller PD can be and still have (D, W) = (2, 6) as the optimal solution. In other words, how much can the estimate of $300 for the unit profit for these doors be off before the model will give an erroneous optimal solution? Using the Spreadsheet to Do Sensitivity Analysis One of the great strengths of a spreadsheet is the ease with which it can be used interactively to perform various kinds of what-if analysis, including the sensitivity analysis being considered 5.3 The Effect of Changes in One Objective Function Coefficient 157 Run Solver again and the spreadsheet immediately reveals the effect of changing any values in the data cells. in this section. Once Solver has been set up to obtain an optimal solution, you can immediately find out what would happen if one of the parameters of the model were to be changed to some other value. All you have to do is make this change on the spreadsheet and then run Solver again. To illustrate, Figure 5.2 shows what would happen if the unit profit for doors were to be decreased from PD = $300 to PD = $200. Comparing with Figure 5.1, there is no change at all in the optimal solution. In fact, the only changes in the new spreadsheet are the new value of PD in cell C4 and a decrease of $200 in the total profit shown in cell G12 (because each of the two doors produced per week provides $100 less profit). Because the optimal solution does not change, we now know that the original estimate of PD = $300 can be considerably too high without invalidating the model’s optimal solution. But what happens if this estimate is too low instead? Figure 5.3 shows what would happen if PD were to be increased to PD = $500. Again, there is no change in the optimal solution. Because the original value of PD = $300 can be changed considerably in either direction without changing the optimal solution, PD is said to be not a sensitive parameter. It is not necessary to pin down this estimate with great accuracy to have confidence that the model is providing the correct optimal solution. This may be all the information that is needed about PD. However, if there is a good ­possibility that the true value of PD will turn out to be outside this broad range from $200 to $500, further investigation would be desirable. How much higher or lower can PD be before the optimal solution would change? FIGURE 5.2 The revised Wyndor problem where the estimate of the unit profit for doors has been decreased from PD = $300 to PD = $200, but no change occurs in the optimal solution. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 Unit Profit 4 Doors Windows $200 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 2 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 2 6 $3,400 C D 10 11 12 FIGURE 5.3 The revised Wyndor problem where the estimate of the unit profit for doors has been increased from PD = $300 to PD = $500, but no change occurs in the optimal solution. Units Produced A 1 B E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $500 $500 5 6 7 Hours Used per Unit Produced Plant 1 1 0 Hours Hours Used Available 2 ≤ 4 12 18 8 Plant 2 0 2 12 ≤ 9 Plant 3 3 2 18 ≤ Doors Windows Total Profit 2 6 $4,000 10 11 12 Units Produced 158 Chapter Five What-If Analysis for Linear Programming Figure 5.4 demonstrates that the optimal solution would indeed change if PD were increased all the way up to PD = $1,000. Thus, we now know that this change occurs somewhere between $500 and $1,000 during the process of increasing PD. Using a Parameter Analysis Report Generated by Analytic Solver to Do Sensitivity Analysis Systematically To pin down just when the optimal solution will change, we could continue selecting new values of PD at random. However, a better approach is to systematically consider a range of values of PD. The Analytic Solver, first introduced in Section 2.6, can generate a parameter analysis report that is designed to do just this sort of analysis. Instructions for downloading this software are available in the preface and at www.mhhe.com/Hillier6e. In preparation for running a parameter analysis report, several additional range names were defined for specific cells that will be changed (e.g., UnitProfitPerDoor for C4) and for the specific changing cells (DoorsProduced and WindowsProduced for C12 and D12, respectively). This allows the resulting report to show the results with more informative names for each of these cells. The data cell containing a parameter that will be systematically varied (UnitProfitPerDoor, or C4, in this case) is referred to as a parameter cell. A parameter analysis report is used to show the results in the changing cells and/or the objective cell for various trial values in the parameter cell. For each trial value, these results are obtained by using Solver to re-solve the problem. To generate a parameter analysis report, the first step is to define the parameter cell. In this case, select cell C4 (UnitProfitPerDoor) and choose Parameters > Optimization on the Analytic Solver ribbon. In the parameter cell dialog box, shown in Figure 5.5, enter the range FIGURE 5.4 The revised Wyndor problem where the estimate of the unit profit for doors has been increased from PD = $300 to PD = $1,000, which results in a change in the optimal solution. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $1,000 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 4 ≤ 4 8 Plant 2 0 2 6 12 9 Plant 3 3 2 18 ≤ ≤ Doors Windows Total Profit 4 3 $5,500 18 10 11 12 FIGURE 5.5 The parameter cell dialog box for PD specifies here that this parameter cell for the Wyndor problem will be systematically varied from $100 to $1,000. Units Produced 5.3 The Effect of Changes in One Objective Function Coefficient 159 of trial values for the parameter cell. The entries shown specify that PD will be systematically varied from $100 to $1,000. If desired, additional parameter cells could be defined in this same way, but we will not do so at this point. Next choose Reports > Optimization > Parameter Analysis on the Analytic Solver ribbon. This brings up the dialog box shown in Figure 5.6 that allows you to specify which parameter cells to vary and which results to show. The choice of which parameter cells to vary is made under Parameters in the bottom half of the dialog box. Clicking on (>>) will select all of the parameter cells defined so far (moving them to the box on the right). In the ­Wyndor ­example, only one parameter has been defined, so this causes the single parameter cell ­(UnitProfitPerDoor) to appear on the right. If more parameter cells had been defined, particular parameter cells can be chosen for immediate analysis by clicking on the + next to Wyndor to reveal the list of parameter cells that have been defined in the Wyndor spreadsheet. Clicking on (>) then moves individual parameter cells to the list on the right. The choice of which results to show as the parameter cell is varied is made in the upper half of the dialog box. Clicking on (>>) will cause all of the changing cells (DoorsProduced and WindowsProduced) and the objective cell (Total Profit) to appear in the list on the right. To instead choose a subset of these cells, click on the small + next to Variables (or Objective) to reveal a list of all the changing cells (or objective cell) and then click on > to move that changing cell (or objective cell) to the right. Finally, enter the number of Major Axis Points to specify how many different values of the parameter cell will be shown in the parameter analysis report. The values will be spread evenly between the lower and upper values specified in the parameters cell dialog box in ­Figure 5.5. With 10 major axis points, a lower value of $100, and an upper value of $1000, the parameter analysis report will show results for PD of $100, $200, $300, . . ., $1000. FIGURE 5.6 The dialog box for the parameter analysis report specifies here for the Wyndor problem that the UnitProfitPerDoor parameter cell will be varied and that results from all the changing cells (DoorsProduced and WindowsProduced) and the objective cell (Total-Profit) will be shown. 160 Chapter Five What-If Analysis for Linear Programming The allowable range for a coefficient in the objective function is the range of values for this coefficient over which the optimal solution for the original model remains optimal. Clicking on the OK button generates the parameter analysis report shown in Figure 5.7. One at a time, the trial values listed in the first column of the table are put into the parameter cell (UnitProfitPerDoor) and then Solver is called on to re-solve the problem. The optimal results for that particular trial value of the parameter cell are then shown in the remaining columns— DoorsProduced, WindowsProduced, and TotalProfit. This is repeated automatically for each remaining trial value of the parameter cell. The end result (which happens very quickly for small problems) is the completely-filled-in parameter analysis report shown in Figure 5.7. The parameter analysis report reveals that the optimal solution remains the same all the way from PD = $100 (and perhaps lower) to PD = $700, but that a change occurs somewhere between $700 and $800. We next could systematically consider values of PD between $700 and $800 to determine more closely where the optimal solution changes. However, here is a shortcut. The range of values of PD over which (D, W) = (2, 6) remains as the optimal solution is referred to as the allowable range for an objective function coefficient, or just the allowable range for short. Upon request, Solver will provide a report called the sensitivity report that reveals exactly what this allowable range is. Using the Sensitivity Report to Find the Allowable Range As was shown in Figure 2.12, when Excel’s Solver gives the message that it has found a solution, it also gives on the right a list of three reports that can be provided. By selecting the ­second one (labeled Sensitivity), you will obtain the sensitivity report. With Analytic Solver, this same report is obtained by choosing Reports > Optimization > Sensitivity from the A ­ nalytic Solver ribbon. Figure 5.8 shows the relevant part of this report for the Wyndor problem. The Final Value column indicates the optimal solution. The next column gives the reduced costs, which can provide some useful information when any of the changing cells equal zero in the optimal solution, which is not the case here. (For a zero-valued changing cell, the corresponding reduced cost can FIGURE 5.7 The parameter analysis report that shows the effect of systematically varying the estimate of the unit profit for doors for the Wyndor problem. A B C D 1 UnitProfitPerDoor DoorsProduced WindowsProduced TotalProfit 2 $100 2 6 $3,200 3 $200 2 6 $3,400 4 $300 2 6 $3,600 5 $400 2 6 $3,800 6 $500 2 6 $4,000 $4,200 7 $600 2 6 8 $700 2 6 $4,400 9 $800 4 3 $4,700 10 $900 4 3 $5,100 11 $1,000 4 3 $5,500 FIGURE 5.8 Part of the sensitivity report generated by Solver for the original Wyndor problem (Figure 5.1), where the last three columns enable identifying the allowable ranges for the unit profits for doors and windows. Variable Cells Cell Name Final Value Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease $C$12 $D$12 DoorsProduced WindowsProduced 2 6 0 0 300 500 450 IE + 30 300 300 5.3 The Effect of Changes in One Objective Function Coefficient 161 Reduced costs are described in the supplement to this chapter at www.mhhe. com/Hillier6e. Don’t worry about this relatively technical subject (unless your instructor assigns this supplement). The sensitivity report generated by Solver reveals the allowable range for each coefficient in the objective function. be used to determine what the effect would be of either increasing that changing cell or making a change in its coefficient in the objective function. Because of the relatively technical nature of these interpretations of reduced costs, we will not discuss them further here, but will provide a full explanation in the supplement to this chapter at www.mhhe.com/­Hillier6e. The next three columns provide the information needed to identify the allowable range for each coefficient in the objective function. The Objective Coefficient column gives the current value of each coefficient, and then the next two columns give the allowable increase and the allowable decrease from this value to remain within the allowable range. For example, consider PD, the coefficient of D in the objective function. Since D is the production rate for these special doors, the Doors row in the table provides the following information (without the dollar sign) about PD: Current value of P D: 300 Allowable increase in P D: 450 So P D ≤ 300 + 450 = 750 Allowable decrease in P D: 300 So P D ≥ 300 − 300 = 0 Allowable range for P D: 0 ≤ P D ≤ 750 Therefore, if PD is changed from its current value (without making any other change in the model), the current solution (D, W) = (2, 6) will remain optimal so long as the new value of PD is within this allowable range. Figure 5.9 provides graphical insight into this allowable range. For the original value of PD = 300, the solid line in the figure shows the slope of the objective function line passing through (2, 6). At the lower end of the allowable range, when PD = 0, the objective function line that passes through (2, 6) now is line B in the figure, so every point on the line segment between (0, 6) and (2, 6) is an optimal solution. For any value of PD < 0, the objective function line will have rotated even further so that (0, 6) becomes the only optimal solution. At the upper end of the allowable range, when PD = 750, the objective function line that passes through (2, 6) becomes line C, so every point on the line segment between (2, 6) and (4, 3) becomes an optimal solution. For any value of PD > 750, the objective function line is even steeper than line C, so (4, 3) becomes the only optimal solution. Conclusion: The allowable range for PD is 0 ≤ PD ≤ 750, because (D, W) = (2, 6) remains optimal over this range but not beyond. (When PD = 0 or PD = 750, there are multiple optimal FIGURE 5.9 The two dashed lines that pass through solid constraint boundary lines are the objective function lines when PD (the unit profit for doors) is at an endpoint of its allowable range, 0 ≤ PD ≤ 750, since either line or any objective function line in between still yields (D, W) = (2, 6) as an optimal solution for the Wyndor problem. Production rate for windows W 8 (2, 6) is optimal for 0 ≤ PD ≤ 750 6 Line B PD = 0 (Profit = 0D + 500W) Line C 4 PD = 300 (Profit = 300D + 500W) Feasible region 2 PD = 750 (Profit = 750D + 500W) Line A 0 2 4 6 Production rate for doors D 162 Chapter Five What-If Analysis for Linear Programming solutions, but (D, W) = (2, 6) still is one of them.) With the range this wide around the original estimate of $300 (PD = 300) for the unit profit for doors, we can be quite confident of obtaining the correct optimal solution for the true unit profit even though the discussion in Section 5.2 indicates that this estimate is fairly rough. The sensitivity report also can be used to find the allowable range for the unit profit for Wyndor’s other new product. In particular, let P W = Unit profit for Wyndor’s new kind of window = Cell D4 in the spreadsheet Referring to the Windows row of the sensitivity report (Figure 5.8), this row indicates that the allowable decrease in PW is 300 (so PW ≥ 500 − 300 = 200) and the allowable increase is 1E + 30. What is meant by 1E + 30? This is shorthand in Excel for 1030 (1 with 30 zeroes after it). This tremendously huge number is used by Excel to represent infinity. Therefore, the allowable range of PW is obtained from the sensitivity report as follows: Current value of P W: 500 Allowable increase in P W: Unlimited So P W has no upper limit Allowable decrease in P W: 300 So P W ≥ 500 − 300 = 200 Allowable range: P W ≥ 200 A parameter is considered sensitive if even a small change in its value can change the optimal solution. Review Questions The allowable range is quite wide for both objective function coefficients. Thus, even though PD = $300 and PW = $500 were only rough estimates of the true unit profit for the doors and windows, respectively, we can still be confident that we have obtained the correct optimal solution. We are not always so lucky. For some linear programming problems, even a small change in the value of certain coefficients in the objective function can change the optimal solution. As first mentioned in Section 5.1, such coefficients are referred to as sensitive parameters. The sensitivity report will immediately indicate which of the objective function coefficients (if any) are sensitive parameters. These are parameters that have a small allowable increase and/or a small allowable decrease. Hence, extra care should be taken to refine these estimates. Once this has been done and the final version of the model has been solved, the allowable ranges continue to serve an important purpose. As indicated in Section 5.1, the second benefit of what-if analysis is that if conditions change after the study has been completed (a common occurrence), what-if analysis leaves signposts that indicate (without solving the model again) whether a resulting change in a parameter of the model changes the optimal solution. Thus, if weeks, months, or even years later, the unit profit for one of Wyndor’s new products changes substantially, its allowable range indicates immediately whether the old optimal product mix still is the appropriate one to use. Being able to draw an affirmative conclusion without reconstructing and solving the revised model is extremely helpful for any linear programming problem, but especially so when the model is a large one. 1. What type of report can Analytic Solver generate to show how the optimal solution varies for a defined range of values of a parameter cell? 2. What is meant by the allowable range for a coefficient in the objective function? 3. What is the significance if the true value for a coefficient in the objective function turns out to be so different from its estimate that it lies outside its allowable range? 4. In Solver’s sensitivity report, what is the interpretation of the Objective Coefficient column? The Allowable Increase column? The Allowable Decrease column? 5.4 THE EFFECT OF SIMULTANEOUS CHANGES IN OBJECTIVE ­FUNCTION COEFFICIENTS The coefficients in the objective function typically represent quantities (e.g., unit profits) that can only be estimated because of considerable uncertainty about what their true values will turn out to be. The allowable ranges described in the preceding section deal with this uncertainty by 5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 163 focusing on just one coefficient at a time. In effect, the allowable range for a particular coefficient assumes that the original estimates for all the other coefficients are completely accurate so that this coefficient is the only one whose true value may differ from its original estimate. In actuality, the estimates for all the coefficients (or at least more than one of them) may be inaccurate simultaneously. The crucial question is whether this is likely to result in obtaining the wrong optimal solution. If so, greater care should be taken to refine these estimates as much as possible, at least for the more crucial coefficients. On the other hand, if what-if analysis reveals that the anticipated errors in estimating the coefficients are unlikely to affect the optimal solution, then management can be reassured that the current linear programming model and its results are providing appropriate guidance. This section focuses on how to determine, without solving the problem again, whether the optimal solution might change if certain changes occur simultaneously in the coefficients of the objective function (due to their true values differing from their estimates). In the process, we will address the second of Wyndor management’s what-if questions. Question 2: What happens if the estimates of the unit profits of both of Wyndor’s new products are inaccurate? Using the Spreadsheet for This Analysis Once again, a quick-and-easy way to address this kind of question is to simply try out different estimates on the spreadsheet formulation of the model and see what happens each time after running Solver again. In this case, the optimal product mix indicated by the model is heavily weighted toward producing the windows (6 per week) rather than the doors (only 2 per week). Since there is equal enthusiasm for both new products, management is concerned about this imbalance. Therefore, management has raised a what-if question. What would happen if the estimate of the unit profit for the doors ($300) were too low and the corresponding estimate for the windows ($500) were too high? Management feels that the estimates could easily be off in these directions. If this were the case, would this lead to a more balanced product mix being the most profitable one? This question can be answered in a matter of seconds simply by substituting new estimates of the unit profits in the original spreadsheet in Figure 5.1 and running Solver again. ­Figure 5.10 shows that new estimates of $450 for doors and $400 for windows causes no change at all in the solution for the optimal product mix. (The total profit does change, but this occurs only because of the changes in the unit profits.) Would even larger changes in the estimates of unit profits finally lead to a change in the optimal product mix? Figure 5.11 shows that this does happen, yielding a relatively balanced product mix of (D, W) = (4, 3), when estimates of $600 for doors and $300 for windows are used. FIGURE 5.10 The revised Wyndor problem where the estimates of the unit profits for doors and windows have been changed to PD = $450 and PW = $400, respectively, but no change occurs in the optimal solution. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $450 $400 5 6 Hours Used per Unit Produced Hours Hours Used Available ≤ 4 12 ≤ 12 18 ≤ 18 7 Plant 1 1 0 2 8 Plant 2 0 2 9 Plant 3 3 2 Doors Windows Total Profit 2 6 $3,300 10 11 12 Units Produced 164 Chapter Five What-If Analysis for Linear Programming FIGURE 5.11 The revised Wyndor problem where the estimates of the unit profits for doors and windows have been changed to $600 and $300, respectively, which results in a change in the optimal solution. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $600 $300 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 4 ≤ 4 8 Plant 2 0 2 6 ≤ 12 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 4 3 $3,300 10 11 12 Units Produced Using a Two-Way Parameter Analysis Report Generated by Analytic Solver for This Analysis Using Analytic Solver, a two-way parameter analysis report provides a way of systematically investigating the effect if the estimates of both unit profits are inaccurate. This kind of parameter analysis table shows the results in a single output cell for various trial values in two parameter cells. Therefore, for example, it can be used to show how TotalProfit (G12) in Figure 5.1 varies over a range of trial values in the two parameter cells, UnitProfitPerDoor (C4) and UnitProfitPerWindow (D4). For each pair of trial values in these data cells, Solver is called on to re-solve the problem. To create such a two-way parameter analysis report for the Wyndor problem, both UnitProfitPerDoor (C4) and UnitProfitPerWindow (D4) need to be defined as parameter cells. In turn, select cell C4 and D4, then choose Parameters > Optimization on the Analytic Solver ribbon, and then enter the range of trial values for each parameter cell (as was done in ­Figure 5.5 in the preceding section). For this example, UnitProfitPerDoor (C4) will be varied from $300 to $600 while UnitProfitPerWindow (D4) will be varied from $100 to $500. Next, choose Reports > Optimization > Parameter Analysis on the Analytic Solver ribbon to bring up the dialog box shown in Figure 5.12. For a two-way parameter analysis report, two parameter cells are chosen, but only a single result can be shown. Under Parameters, clicking on (>>) chooses both of the defined parameter cells, UnitProfitPerDoor and UnitProfitPer­Window. Under Results, click on (<<) to clear out the list of cells on the right, click on the + next to Objective to reveal the objective cell (TotalProfit), select TotalProfit, and then click on > to move this cell to the right. The next step is to choose Vary Two Selected Parameters Independently from the menu toward the bottom of the window (as opposed to the other two options—Vary All Selected Parameters Simultaneously or One At a Time). This will allow both parameter cells to be varied independently over their entire ranges. The number of different values of the first parameter cell and the second parameter cell to be shown in the parameter analysis report are entered in Major Axis Points and Minor Axis Points, respectively. These values will be spread evenly over the range of values specified in the parameter dialog box for each parameter cell. Therefore, choosing 4 and 5 for the respective number of values, as shown in Figure 5.12, will vary UnitProfitPerDoor over the four values of $300, $400, $500, and $600 while simultaneously varying UnitProfitPerWindow over the five values of $100, $200, $300, $400, and $500. Clicking on the OK button generates the parameter analysis report shown in Figure 5.13. The trial values for the respective parameter cells are listed in the first column and first row of the table. For each combination of a trial value from the first column and from the first row, Solver has solved for the value of the output cell of interest (the objective cell for this example) and entered it into the corresponding column and row of the table. 5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 165 FIGURE 5.12 The dialog box for the parameter analysis report specifies here that the UnitProfitPerDoor and UnitProfitPerWindow parameter cells will be varied and results from the objective cell (TotalProfit) will be shown for the Wyndor problem. FIGURE 5.13 The parameter analysis report that shows how the optimal total profit changes when systematically varying the estimate of both the unit profit for doors and the unit profit for windows for the Wyndor problem. What-if analysis shows that there is no need to refine Wyndor’s estimates of the unit profits for doors and windows. 1 2 3 4 5 6 A TotalProfit UnitProfitPerDoor $300 $400 $500 $600 B UnitProfitPerWindow $100 $1,500 $1,900 $2,300 $2,700 C D E $200 $1,800 $2,200 $2,600 $3,000 $300 $2,400 $2,600 $2,900 $3,300 $400 $3,000 $3,200 $3,400 $3,600 F $500 $3,600 $3,800 $4,000 $4,200 It also is possible to choose either DoorsProduced or WindowsProduced instead of TotalProfit as the result to show in the dialog box in Figure 5.12. A similar parameter analysis report then could have been generated to show either the optimal number of doors to produce or the optimal number of windows to produce for each combination of values for the unit profits. These two parameter analysis reports are shown in Figure 5.14. The upper right-hand corner (cell F3) of both reports, taken together, gives the optimal solution of (D, W) = (2, 6) when using the original unit-profit estimates of $300 for doors and $500 for windows. ­Moving down from this cell corresponds to increasing this profit estimate for doors, while moving to the left amounts to decreasing the profit estimate for windows. Moving to the lower-left corner of both tables reveals that (D, W) = (4, 3) becomes the optimal solution for these values of the profit estimates. However, (D, W) = (2, 6) continues to be the optimal solution for all the cells near F3. This indicates that the original estimates of unit profit would need to be very inaccurate indeed before the optimal product mix would change. Although the estimates are fairly rough, management is confident that they are not that inaccurate. Therefore, there is no need to expend the considerable effort that would be needed to refine the estimates. 166 Chapter Five What-If Analysis for Linear Programming FIGURE 5.14 The pair of parameter analysis reports that show how the optimal number of doors to produce (top report) and the optimal number of windows to produce (bottom report) change when systematically varying the estimate of both the unit profit for doors and the unit profit for windows for the Wyndor problem. 1 2 3 4 5 6 A DoorsProduced UnitProfitPerDoor $300 $400 $500 $600 B UnitProfitPerWindow $100 4 4 4 4 1 2 3 4 5 6 A WindowsProduced UnitProfitPerDoor $300 $400 $500 $600 B UnitProfitPerWindow $100 3 3 3 3 C D E $200 2 4 4 4 $300 2 2 4 4 $400 2 2 2 2 C D E F $200 6 3 3 3 $300 6 6 3 3 $400 6 6 6 6 $500 6 6 6 6 F $500 2 2 2 2 At this point, it continues to appear that (D, W) = (2, 6) is the best product mix for initiating the production of the two new products (although additional what-if questions remain to be addressed in subsequent sections). However, we also now know from Figure 5.14 that as conditions change in the future, if the unit profits for both products change enough, it may be advisable to change the product mix later. We still need to leave clear signposts behind to signal when a future change in the product mix should be considered, as described next. Gleaning Additional Information from the Sensitivity Report The preceding section described how the data in the sensitivity report enable finding the allowable range for an individual coefficient in the objective function when that coefficient is the only one that changes from its original value. These same data (the allowable increase and allowable decrease in each coefficient) also can be used to analyze the effect of simultaneous changes in these coefficients. Here is how. A sum ≤ 100 percent guarantees that the original optimal solution is still optimal. The 100 Percent Rule for Simultaneous Changes in Objective Function Coefficients: If simultaneous changes are made in the coefficients of the objective function, calculate for each change the percentage of the allowable change (increase or decrease) for that coefficient to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the original optimal solution definitely will still be optimal. (If the sum does exceed 100 percent, then we cannot be sure.) This rule does not spell out what happens if the sum of the percentage changes does exceed 100 percent. The consequence depends on the directions of the changes in the coefficients. Exceeding 100 percent may or may not change the optimal solution, but so long as 100 percent is not exceeded, the original optimal solution definitely will still be optimal. Keep in mind that we can safely use the entire allowable increase or decrease in a single objective function coefficient only if none of the other coefficients have changed at all. With simultaneous changes in the coefficients, we focus on the percentage of the allowable increase or decrease that is being used for each coefficient. To illustrate, consider the Wyndor problem again, along with the information provided by the sensitivity report in Figure 5.8. Suppose conditions have changed after the initial study, and the unit profit for doors (PD) has increased from $300 to $450 while the unit profit for windows (PW) has decreased from $500 to $400. The calculations for the 100 percent rule then are P D: $300 → $450 450 − 300 Percentage of allowable increase = 100 ____________ % = 331⁄3% ( 450 ) P W : $500 → $400 500 − 400 1 ____________ Percentage of allowable decrease = 100 % = 33⁄3% ( 300 ) ¯ Sum = 66 2⁄3% 5.4 The Effect of Simultaneous Changes in Objective ­Function Coefficients 167 Since the sum of the percentages does not exceed 100 percent, the original optimal solution (D, W) = (2, 6) definitely is still optimal, just as we found earlier in Figure 5.10. Now suppose conditions have changed even further, so PD has increased from $300 to $600 while PW has decreased from $500 to $300. The calculations for the 100 percent rule now are P D: $300 → $600 600 − 300 2 Percentage of allowable increase = 100 ____________ ( 450 )% = 66⁄3% P W : $500 → $300 500 − 300 Percentage of allowable decrease = 100 ____________ % = 662⁄3% ( 300 ) ¯ Sum = 133 1⁄3% Since the sum of the percentages now exceeds 100 percent, the 100 percent rule says that we can no longer guarantee that (D, W) = (2, 6) is still optimal. In fact, we found earlier in both Figures 5.11 and 5.14 that the optimal solution has changed to (D, W) = (4, 3). These results suggest how to find just where the optimal solution changes while PD is being increased and PW is being decreased in this way. Since 100 percent is midway between 66 2⁄3 percent and 1331⁄3percent, the sum of the percentage changes will equal 100 percent when the values of PD and PW are midway between their values in the above cases. In particular, PD = $525 is midway between $450 and $600 and PW = $350 is midway between $400 and $300. The corresponding calculations for the 100 percent rule are P D: $300 → $525 525 − 300 Percentage of allowable increase = 100 ____________ % = 50% ( 450 ) P W: $500 → $300 500 − 350 Percentage of allowable decrease = 100 ____________ % = 50% ( 300 ) Sum = ¯ 100% Although the sum of the percentages equals 100 percent, the fact that it does not exceed 100 percent guarantees that (D, W) = (2, 6) is still optimal. Figure 5.15 shows graphically that both (2, 6) and (4, 3) are now optimal, as well as all the points on the line segment connecting these two points. However, if PD and PW were to be changed any further from their original values (so that the sum of the percentages exceeds 100 percent), the objective function line would be rotated so far toward the vertical that (D, W) = (4, 3) would become the only optimal solution. At the same time, keep in mind that having the sum of the percentages of allowable changes exceed 100 percent does not automatically mean that the optimal solution will change. For example, suppose that the estimates of both unit profits are halved. The resulting calculations for the 100 percent rule are P D: $300 → $150 300 − 150 Percentage of allowable decrease = 100 ____________ % = 50% ( 300 ) P W: $500 → $250 500 − 250 Percentage of allowable decrease = 100 ____________ % = 83% ( 300 ) Sum = ¯ 133% Here is an example where the original optimal solution is still optimal even though the sum exceeds 100 percent. Even though this sum exceeds 100 percent, Figure 5.16 shows that the original optimal solution is still optimal. In fact, the objective function line has the same slope as the original objective function line (the solid line in Figure 5.9). This happens whenever proportional changes are made to all the unit profits, which will automatically lead to the same optimal solution. 168 Chapter Five What-If Analysis for Linear Programming FIGURE 5.15 When the estimates of the unit profits for doors and windows change to PD = $525 and PW = $350, which lies at the edge of what is allowed by the 100 percent rule, the graphical method shows that (D, W) = (2, 6) still is an optimal solution, but now every other point on the line segment between this solution and (4, 3) also is optimal. Production rate for windows W 10 8 Objective function line now is Profit = $3,150 = 525D + 350W since PD = $525 and PW = $350 (2, 6) 6 Entire line segment is optimal 4 Feasible region (4, 3) 2 0 FIGURE 5.16 When the estimates of the unit profits for doors and windows change to PD = $150 and PW = $250 (half their original values), the graphical method shows that the optimal solution still is (D, W) = (2, 6), even though the 100 percent rule says that the optimal solution might change. 2 4 Production rate for doors Production rate for windows Profit = $1,800 = 150D + 250W 6 8 D W 8 Optimal solution 6 (2, 6) 4 Feasible region 2 0 2 4 Production rate for doors 6 8 D Comparisons You now have seen three approaches to investigating what happens if simultaneous changes occur in the coefficients of the objective function: (1) try out changes directly on a spreadsheet, (2) use a two-way parameter analysis report and (3) apply the 100 percent rule. 5.5 The Effect of Single Changes in a Constraint 169 The spreadsheet approach is a good place to start, especially for less experienced modelers, because it is simple and quick. If you are only interested in checking one specific set of changes in the coefficients, you can immediately see what happens after making the changes in the spreadsheet. More often, there will be numerous possibilities for what the true values of the coefficients will turn out to be, because of uncertainty in the original estimates of these coefficients. The parameter analysis report is useful for systematically checking a variety of possible changes in one or two objective function coefficients. Trying out representative possibilities on the spreadsheet may provide all the insight that is needed. Perhaps the optimal solution for the original model will remain optimal over nearly all these possibilities, so this solution can be confidently used. Or perhaps it will become clear that the original estimates need to be refined before selecting a solution. When the spreadsheet approach and/or parameter analysis report does not provide a clear conclusion, the 100 percent rule can usefully complement this approach in the following ways: • The 100 percent rule can be used to determine just how large the changes in the objective function coefficients need to be before the original optimal solution may no longer be optimal. • When the model has a large number of decision variables (as is common for real problems), it may become impractical to use the spreadsheet approach to systematically try out a variety of simultaneous changes in many or all of the coefficients in the objective function because of the huge number of representative possibilities. The parameter analysis report can only be used to systematically check possible changes in—at most—two coefficients at a time. However, by dividing each coefficient’s allowable increase or allowable decrease by the number of decision variables, the 100 percent rule immediately indicates how much each coefficient can be safely changed without invalidating the current optimal solution. • After completing the study, if conditions change in the future that cause some or all of the coefficients in the objective function to change, the 100 percent rule quickly indicates whether the original optimal solution must remain optimal. If the answer is affirmative, there is no need to take all the time that may be required to reconstruct the (revised) spreadsheet model. The time saved can be very substantial for large models. Review Questions 5.5 1. How many result cells can be selected for display in Analytic Solver’s two-way parameter analysis report? 2. In the 100 percent rule for simultaneous changes in objective function coefficients, what are the percentage changes that are being considered? 3. In this 100 percent rule, if the sum of the percentage changes does not exceed 100 percent, what does this say about the original optimal solution? 4. In this 100 percent rule, if the sum of the percentage changes exceeds 100 percent, does this mean that the original optimal solution is no longer optimal? THE EFFECT OF SINGLE CHANGES IN A CONSTRAINT When the right-hand sides represent managerial policy decisions, what-if analysis provides guidance regarding the effect of altering these decisions. We now turn our focus from the coefficients in the objective function to the effect of changing the functional constraints. The changes might occur either in the coefficients on the left-hand sides of the constraints or in the values of the right-hand sides. We might be interested in the effect of such changes for the same reason we are interested in this effect for objective function coefficients, namely, that these parameters of the model are only estimates of quantities that cannot be determined precisely at this time so we want to determine the effect if these estimates are inaccurate. However, a more common reason for this interest is the one discussed at the end of ­Section 5.1, namely, that the right-hand sides of the functional constraints may well represent managerial policy decisions rather than quantities that are largely outside the control of management. Therefore, after the model has been solved, management will want to analyze the effect of altering these policy decisions in a variety of ways to see if these decisions can 170 Chapter Five What-If Analysis for Linear Programming be improved. What-if analysis provides valuable guidance to management in determining the effect of altering these policy decisions. (Recall that this was cited as the third benefit of what-if analysis in Section 5.1.) This section describes how to perform what-if analysis when making changes in just one spot (a coefficient or a right-hand side) of a single constraint. The next section then will deal with simultaneous changes in the constraints. The procedure for determining the effect if a single change is made in a constraint is the same regardless of whether the change is in a coefficient on the left-hand side or in the value on the right-hand side. (The one exception is that the Solver sensitivity report provides information about changes in the right-hand side but does not do so for the left-hand side.) Therefore, we will illustrate the procedure by making changes in a right-hand side. In particular, we return to the Wyndor case study to address the third what-if question posed by Wyndor management in Section 5.2. Question 3: What happens if a change is made in the number of hours of production time per week being made available to Wyndor’s new products in one of the plants? The number of hours available in each plant is the value of the right-hand side for the corresponding constraint, so we want to investigate the effect of changing this right-hand side for one of the plants. With the original optimal solution, (D, W) = (2, 6), only 2 of the 4 available hours in Plant 1 are used, so changing this number of available hours (barring a large decrease) would have no effect on either the optimal solution or the resulting total profit from the two new products. However, it is unclear what would happen if the number of available hours in either Plant 2 or Plant 3 were to be changed. Let’s start with Plant 2. Using the Spreadsheet for This Analysis Referring back to Section 5.2, Figure 5.1 shows the spreadsheet model for the original Wyndor problem before beginning what-if analysis. The optimal solution is (D, W) = (2, 6) with a total profit of $3,600 per week from the two new products. Cell G8 shows that 12 hours of production time per week are being made available for the new products in Plant 2. To see what happens if a specific change is made in this number of hours, all you need to do is substitute the new number in cell G8 and run Solver again. For example, Figure 5.17 shows the result if the number of hours is increased from 12 to 13. The corresponding optimal solution in C12:D12 gives a total profit of $3,750. Thus, the resulting change in profit would be Incremental profit = $3,750 − $3,600 = $150 Since this increase in profit is obtained by adding just one more hour in Plant 2, it would be interesting to see the effect of adding several more hours. Figure 5.18 shows the effect of FIGURE 5.17 The revised Wyndor problem where the hours available in Plant 2 per week have been increased from 12 (as in Figure 5.1) to 13, which results in an increase of $150 in the total profit per week from the two new products. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 1.66667 ≤ 4 8 Plant 2 0 2 13 ≤ 13 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 1.667 6.5 $3,750 10 11 12 Units Produced 5.5 The Effect of Single Changes in a Constraint 171 FIGURE 5.18 A further revision of the Wyndor problem in Figure 5.17 to further increase the hours available in Plant 2 from 13 to 18, which results in a further increase in total profit of $750 (which is the $150 per hour added in Plant 2). A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 Unit Profit 4 Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0 ≤ 4 8 Plant 2 0 2 18 ≤ 18 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 0 9 $4,500 10 11 12 Units Produced adding five more hours. Comparing Figure 5.18 to Figure 5.17, the additional profit from providing five more hours would be So far, each additional hour provided in Plant 2 adds $150 to profit. Now management needs to consider the trade-off between adding production time for the new products and decreasing it for other products. Incremental profit = $4,500 − $3,750 = $750 from adding 5 hours = $150 per hour added Would adding even more hours increase profit even further? Figure 5.19 shows what would happen if a total of 20 hours per week were made available to the new products in Plant 2. Both the optimal solution and the total profit are the same as in Figure 5.18, so increasing from 18 to 20 hours would not help. (The reason is that the 18 hours available in Plant 3 prevent producing more than 9 windows per week, so only 18 hours can be used in Plant 2.) Thus, it appears that 18 hours is the maximum that should be considered for Plant 2. However, the fact that the total profit from the two new products can be increased substantially by increasing the number of hours per week made available to the new products from 12 to 18 does not mean that these additional hours should be provided automatically. The production time made available for these two new products can be increased only if it is decreased for other products. Therefore, management will need to assess the disadvantages of decreasing the production time for any other products (including both lost profit and less tangible disadvantages) before deciding whether to increase the production time for the new products. This analysis also might lead to decreasing the production time made available to the two new products in one or more of the plants. FIGURE 5.19 A further revision of the Wyndor problem in Figure 5.18 to further increase the hours available in Plant 2 from 18 to 20, which results in no change in total profit because the optimal solution cannot make use of these additional hours. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0 ≤ 4 8 Plant 2 0 2 18 ≤ 20 9 Plant 3 3 2 18 ≤ 18 Doors Windows Total Profit 0 9 $4,500 10 11 12 Units Produced 172 Chapter Five What-If Analysis for Linear Programming Using a Parameter Analysis Report Generated by Analytic Solver for This Analysis In general, the shadow price for a constraint reveals the rate at which the objective cell can be increased by increasing the right-hand side of that constraint. This remains valid as long as the right-hand side is within its allowable range. This allowable range focuses on a right-hand side and the corresponding shadow price. In contrast to the allowable range for objective function coefficients described in Section 5.3, it does not indicate whether the original solution is still optimal, just whether the shadow price remains valid. Using Analytic Solver, a parameter analysis report can be used to systematically determine the effect of making various changes in one of the parameters in a constraint. For example, Figure 5.20 displays a parameter analysis report (obtained by following the same procedure used to generate a parameter analysis report in Section 5.3) to show how the changing cells and total profit change as the number of available hours in Plant 2 range between 4 and 20. An interesting pattern is apparent in the incremental profit column. Starting at 12 hours available at Plant 2 (the current allotment), each additional hour allocated yields an additional $150 in profit (up to making 18 hours available). Similarly, if hours are taken away from Plant 2, each hour lost causes a loss of $150 profit (down to making six hours available). This rate of change in the profit for increases or decreases in the right-hand side of a constraint is known as the shadow price. Given an optimal solution and the corresponding value of the objective function for a linear programming model, the shadow price for a functional constraint is the rate at which the value of the objective function can be increased by increasing the right-hand side of the constraint by a small amount. However, the shadow price of $150 for the Plant 2 constraint is valid only within a range of values near 12 (in particular, between 6 hours and 18 hours). If the number of available hours is increased beyond 18 hours, then the incremental profit drops to zero. If the available hours are reduced below six hours, then profit drops at a faster rate of $250 per hour. Therefore, letting RHS denote the value of the right-hand side, the shadow price of $150 is valid for 6 ≤ RHS ≤ 18 This range is known as the allowable range for the right-hand side (or just allowable range for short). The allowable range for the right-hand side of a functional constraint is the range of values for this right-hand side over which this constraint’s shadow price remains valid. Unlike the allowable range for objective function coefficients described in Section 5.3, a change that is within the allowable range for the right-hand side does not mean the original FIGURE 5.20 The parameter analysis report that shows the effect of varying the number of hours of production time being made available per week in Plant 2 for Wyndor’s new products. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 A HoursAvailableInPlant2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B DoorsProduced 4 4 4 3.666666667 3.333333333 3 2.666666667 2.333333333 2 1.666666667 1.333333333 1 0.666666667 0.333333333 0 0 0 C WindowsProduced 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9 9 D TotalProfit $2,200 $2,450 $2,700 $2,850 $3,000 $3,150 $3,300 $3,450 $3,600 $3,750 $3,900 $4,050 $4,200 $4,350 $4,500 $4,500 $4,500 E Incremental Profit $250 $250 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $0 $0 5.5 The Effect of Single Changes in a Constraint 173 solution is still optimal. In fact, any time the shadow price is not zero, a change to a right-hand side leads to a change in the optimal solution. The shadow price indicates how much the value of the objective function will change as the optimal solution changes. Using the Sensitivity Report to Obtain the Key Information The shadow prices reveal the relationship between profit and the amount of production time made available in the plants. FIGURE 5.21 The complete sensitivity report generated by Solver for the original Wyndor problem as formulated in Figure 5.1. As illustrated earlier, it is straightforward to use a parameter analysis report to calculate the shadow price for a functional constraint, as well as to find (or at least closely approximate) the allowable range for the right-hand side of this constraint over which the shadow price remains valid. However, this same information also can be obtained immediately from Solver’s sensitivity report for all the functional constraints. Figure 5.21 shows the full sensitivity report provided by Solver for the original Wyndor problem after obtaining the optimal solution given in Figure 5.1. The top half is the part already shown in Figure 5.8 for finding allowable ranges for the objective function coefficients. The bottom half focuses on the functional constraints, including providing the shadow prices for these constraints in the fourth column. The first three columns remind us that (1) the output cells for these constraints in Figure 5.1 are cells E7 to E9, (2) these cells give the number of production hours used per week in the three plants, and (3) the final values in these cells are 2, 12, and 18 (as shown in column E of Figure 5.1). (We will discuss the last three columns a little later.) The shadow price given in the fourth column for each constraint tells us how much the value of the objective function [objective cell (G12) in Figure 5.1] would increase if the righthand side of that constraint (cell G7, G8, or G9) were to be increased by 1. Conversely, it also tells us how much the value of the objective function would decrease if the right-hand side were to be decreased by 1. The shadow price for the Plant 1 constraint is 0, because this plant already is using less hours (2) than are available (4) so there would be no benefit to making an additional hour available. However, Plants 2 and 3 are using all the hours available to them for the two new products (with the product mix given by the changing cells). Thus, it is not surprising that the shadow prices indicate that the objective cell would increase if the hours available in either Plant 2 or Plant 3 were to be increased. To express this information in the language of management, the value of the objective function for this problem [objective cell (G12) in Figure 5.1] represents the total profit in dollars per week from the two new products under consideration. The right-hand side of each functional constraint represents the number of hours of production time being made available per week for these products in the plant that corresponds to this constraint. Therefore, the shadow price for a functional constraint informs management as to how much the total profit from the two new products could be increased for each additional hour of production time made available to these products per week in the corresponding plant. Conversely, the shadow price indicates how much this profit would decrease for each reduction of an hour of production time in that plant. This interpretation of the shadow price remains valid as long as the change in the number of hours of production time is not very large. Specifically, this interpretation of the shadow price remains valid as long as the number of hours of production time remains within its allowable range. Solver’s sensitivity report Variable Cells Name Final Value DoorsProduced WindowsProduced 2 6 Cell $C$12 $D$12 Reduced Cost 0 0 Objective Coefficient 300 500 Allowable Increase 450 IE + 30 Allowable Decrease 300 300 Constraints Cell Name Final Value $E$7 $E$8 $E$9 Plant 1 Used Plant 2 Used Plant 3 Used 2 12 18 Shadow Price 0 150 100 Constraint R. H. Side 4 12 18 Allowable Increase Allowable Decrease IE + 30 6 6 2 6 6 174 Chapter Five What-If Analysis for Linear Programming Here is how to find the allowable ranges for the right-hand sides from the sensitivity report. provides all the data needed to identify the allowable range of each functional constraint. Refer back to the bottom of this report given in Figure 5.21. The final three columns enable calculation of this range. The “Constraint R. H. Side” column indicates the original value of the right-hand side before any change is made. Adding the number in the “Allowable Increase” column to this original value then gives the upper endpoint of the allowable range. Similarly, subtracting the number in the “Allowable Decrease” column from this original value gives the lower endpoint. Using the fact that 1E + 30 represents infinity (∞), these calculations of the allowable ranges are shown below, where a subscript has been added to each RHS to identify the constraint involved. Plant 1 constraint: 4 − 2 ≤ RHS1 ≤ 4 + ∞, so 2 ≤ RHS1 (no upper limit) Plant 2 constraint: 12 − 6 ≤ RHS2 ≤ 12 + 6, so 6 ≤ RHS2 ≤ 18 Plant 3 constraint: 18 − 6 ≤ RHS3 ≤ 18 + 6, so 12 ≤ RHS3 ≤ 24 In the case of the Plant 2 constraint, Figure 5.22 provides graphical insight into why 6 ≤ RHS ≤ 18 is the range of validity for the shadow price. The optimal solution for the original problem, (D, W) = (2, 6), lies at the intersection of line B and line C. The equation for line B is 2W = 12 because this is the constraint boundary line for the Plant 2 constraint (2W ≤ 12). However, if the value of this right-hand side (RHS2 = 12) is changed, line B will either shift upward (for a larger value of RHS2) or downward (for a smaller value of RHS2). As line B shifts, the boundary of the feasible region shifts accordingly and the optimal solution continues to lie at the intersection of the shifted line B and line C—provided the shift in line B is not so large that this intersection is no longer feasible. Each time RHS2 is increased (or decreased) by 1, this intersection shifts enough to increase (or decrease) Profit by the amount of the shadow price ($150). Figure 5.22 indicates that this intersection remains feasible (and so optimal) as RHS2 increases from 12 to 18, because the feasible region expands upward as line B shifts upward. However, for values of RHS2 larger than 18, this intersection is no longer feasible because it gives a negative value of D (the production rate for doors). Thus, each increase of 1 above 18 no longer increases Profit by the amount FIGURE 5.22 A graphical interpretation of the allowable range, 6 ≤ RHS2 ≤ 18, for the right-hand side of Wyndor’s Plant 2 constraint. W Production rate for windows 10 (0, 9) 2W = 18 Profit = 300(0) + 500(9) = $4,500 2W = 12 Profit = 300(2) + 500(6) = $3,600 8 6 4 2 Line B (2, 6) Feasible region for original problem (4, 3) 2W = 6 Profit = 300(4) + 500(3) = $2,700 Line C (3D + 2W = 18) Line A (D = 4) 0 2 4 6 Production rate for doors 8 D 5.6 The Effect of Simultaneous Changes in the Constraints 175 of the shadow price. Similarly, as RHS2 decreases from 12 to 6, this intersection remains feasible (and so optimal) as line B shifts down accordingly. However, for values of RHS2 less than 6, this intersection is no longer feasible because it violates the Plant 1 constraint (D ≤ 4) whose boundary line is line A. Hence, each decrease of 1 below 6 no longer decreases Profit by the amount of the shadow price. Consequently, 6 ≤ RHS ≤ 18 is the allowable range over which the shadow price is valid. Summary Recall again that the right-hand side of each functional constraint for the Wyndor problem represents the number of hours of production time per week in the corresponding plant that is being made available to the two new products. The shadow price for each constraint reveals how much the total profit from these new products would increase for each additional hour of production time made available in the corresponding plant for these products. This interpretation of the shadow price remains valid as long as the number of hours remains within its allowable range. Therefore, each shadow price can be applied by management to evaluate a change in its original decision regarding the number of hours as long as the new number is within the corresponding allowable range. This evaluation also would need to take into account how the change in the number of hours made available to the new products would impact the production rates and profits for the company’s other products. Review Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 5.6 Why might it be of interest to investigate the effect of making changes in a functional constraint? Why might it be possible to alter the right-hand side of a functional constraint? What is meant by a shadow price? How can a shadow price be found by using the spreadsheet? By using a parameter analysis report? By using Solver’s sensitivity report? Why are shadow prices of interest to managers? Can shadow prices be used to determine the effect of decreasing rather than increasing the right-hand side of a functional constraint? What does a shadow price of 0 tell a manager? Which columns of Solver’s sensitivity report are used to find the allowable range for the righthand side of a functional constraint? Why are these allowable ranges of interest to managers? THE EFFECT OF SIMULTANEOUS CHANGES IN THE CONSTRAINTS Managerial policy decisions involving right-hand sides frequently are interrelated, so changes in these decisions should be considered simultaneously. The preceding section described how to perform what-if analysis to investigate the effect of changes in a single spot of a constraint. We now turn our consideration to the effect of simultaneous changes in the constraints. The need to consider these simultaneous changes arises frequently. There may be considerable uncertainty about the estimates for a number of the parameters in the functional constraints, so questions will arise as to the effect if the true values of the parameters simultaneously deviate significantly from the estimates. Since the right-hand sides of the constraints often represent managerial policy decisions, questions will arise about what would happen if some of these decisions were to be changed. These decisions frequently are interrelated and so need to be considered simultaneously. We outline next how the usual three methods for performing what-if analysis can be applied to considering simultaneous changes in the constraints. The third one (using Solver’s sensitivity report) is only helpful for changing right-hand sides. For the first two (using the spreadsheet and using a parameter analysis report), the procedure is the same regardless of whether the changes are in the coefficients on the left-hand sides or in the right-hand sides of the constraints (or both). Since changing the right-hand sides is the more important case, we will focus on this case. In particular, we now will deal with the fourth of Wyndor management’s what-if questions. Question 4: What happens if simultaneous changes are made in the number of hours of production time per week being made available to Wyndor’s new products in all the plants? 176 Chapter Five What-If Analysis for Linear Programming In particular, after seeing that the Plant 2 constraint has the largest shadow price (150), versus a shadow price of 100 for the Plant 3 constraint, management now is interested in exploring a specific type of simultaneous change in these production hours. By shifting the production of one of the company’s current products from Plant 2 to Plant 3, it is possible to increase the number of production hours available to the new products in Plant 2 by decreasing the number of production hours available in Plant 3 by the same amount. Management wonders what would happen if these simultaneous changes in production hours were made. Using the Spreadsheet for This Analysis According to the shadow prices, the effect of shifting one hour of production time per week from Plant 3 to Plant 2 would be as follows. RHS 2: 12 → 13 Change in total profit = Shadow price = $150 100 RHS 3: 18 → 17 Change in total profit = − Shadow price = −_ Net increase in total profit = $50 We now are checking to see whether the shadow prices remain valid for evaluating specific simultaneous changes in the right-hand sides. However, we don’t know if these shadow prices remain valid if both right-hand sides are changed by this amount. A quick way to check this is to substitute the new right-hand sides into the original spreadsheet in Figure 5.1 and run Solver again. The resulting spreadsheet in Figure 5.23 shows that the net increase in total profit (from $3,600 to $3,650) is indeed $50, so the shadow prices are valid for these particular simultaneous changes in right-hand sides. How long will these shadow prices remain valid if we continue shifting production hours from Plant 3 to Plant 2? We could continue checking this by substituting other combinations of right-hand sides into the spreadsheet and re-solving each time. However, a more systematic way of doing this is to use a parameter analysis report, as described next. Using a Parameter Analysis Report Generated by Analytic Solver for This Analysis Since it can become tedious, or even impractical, to use the spreadsheet to investigate a large number of simultaneous changes in the right-hand sides, let us see how Analytic Solver’s parameter analysis report can be used to do this analysis more systematically. We could use a two-way parameter analysis report to investigate how the profit and optimal production rates vary for different combinations of the number of hours available in Plant 2 and Plant 3. However, in this case we aren’t interested in all combinations of hours in the two plants, but rather only those combinations that involve a simple shifting of available hours from Plant 3 to Plant 2. For this analysis, we will see that a one-way parameter analysis table is sufficient. FIGURE 5.23 The revised Wyndor problem where column G in Figure 5.1 has been changed by shifting one of the hours available in Plant 3 to Plant 2 and then re-solving. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 1.33333 4 8 Plant 2 0 2 13 13 9 Plant 3 3 2 17 17 Doors Windows Total Profit 1.333 6.5 $3,650 10 11 12 Units Produced 5.6 The Effect of Simultaneous Changes in the Constraints 177 By entering a formula into one data cell in terms of another one, a one-way parameter analysis report is able to investigate interrelated trial values in both data cells. For each hour reduced in Plant 3, an additional hour is made available in Plant 2. Thus, the number of available hours in Plant 2 is a function of the number of available hours in Plant 3. In particular, since there are 30 total hours available at the two plants (RHS2 + RHS3 = 30), the number of available hours in Plant 2 (RHS2) is RHS 2 = 30 − RHS3 Figure 5.24 shows the Wyndor Glass Co. spreadsheet with the data cell for the number of available hours in Plant 2 replaced by the above formula. Because of this formula, whenever the number of available hours in Plant 3 is reduced, the number of available hours in Plant 2 will automatically increase by the same amount. Now a one-way parameter analysis report can be used to investigate various numbers of available hours in Plant 3 (with the corresponding automatic adjustment made to the available hours at Plant 2). HoursAvailableInPlant3 (G9) is specified as a parameter cell with a range of trial values from 18 down to 12. A parameter analysis report is then generated in Figure 5.25 to show how DoorsProduced (C12), WindowsProduced (D12), and TotalProfit (G12) vary as HoursAvailableInPlant3 (G9) varies from 18 down to 12. A column was added to the left of the parameter analysis report to show how FIGURE 5.24 By inserting a formula into cell G8 that keeps the total number of hours available in Plant 2 and Plant 3 equal to 30, it will be possible to generate a one-way parameter analysis report (see Figure 5.25) that shows the effect of shifting more and more of the hours available from Plant 3 to Plant 2. A 1 B C D E F G H Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 2 4 Total (Plants 2 & 3) 8 Plant 2 0 2 12 12 30 9 Plant 3 3 2 18 18 Doors Windows Total Profit 2 6 $3,600 10 11 12 Units Produced G H 5 Hours 6 Available 7 4 Total (Plants 2 & 3) 8 =H8-G9 30 9 18 FIGURE 5.25 The parameter analysis report that shows the effect of shifting more and more of the hours available from Plant 3 to Plant 2 for the Wyndor problem. 1 2 3 4 5 6 7 8 A HoursAvailableInPlant2 12 13 14 15 16 17 18 B HoursAvailableInPlant3 18 17 16 15 14 13 12 C DoorsProduced 2 1.333 0.667 0 0 0 0 D WindowsProduced 6 6.5 7 7.5 7 6.5 6 E TotalProfit $3,600 $3,650 $3,700 $3,750 $3,500 $3,250 $3,000 F Incremental Profit $50 $50 $50 -$250 -$250 -$250 178 Chapter Five What-If Analysis for Linear Programming HoursAvailableInPlant2 varies correspondingly with the different values of HoursAvailableInPlant3. Also, we have calculated the incremental profit (in column F) for each hour shifted from Plant 3 to Plant 2. Again there is a pattern to the incremental profit. For each hour shifted from Plant 3 to Plant 2 (up to 3 hours), an additional profit of $50 is achieved. However, if more than 3 hours are shifted, the incremental profit per hour shifted becomes −$250. Thus, it appears worthwhile to shift up to 3 available hours from Plant 3 to Plant 2, but no more. Although a one-way parameter analysis report is limited to enumerating trial values for only one data cell, you have just seen how such a parameter analysis report still can systematically investigate a large number of simultaneous changes in two data cells by entering a formula for the second data cell in terms of the first one. The two data cells considered above happened to be right-hand sides of constraints, but either or both could have been coefficients on the left-hand side instead. It is even possible to enter formulas for multiple data cells in terms of the one whose trial values are being enumerated. Furthermore, by using a two-way parameter analysis report, trial values can be enumerated simultaneously for two data cells, with the possibility of entering formulas for additional data cells in terms of these two. Gleaning Additional Information from the Sensitivity Report This 100 percent rule reveals whether the simultaneous changes in the right-hand sides are small enough to guarantee that the shadow prices are still valid. Despite the versatility of a parameter analysis report, it cannot handle a number of important cases. The most important one is where management wants to explore various possibilities for changing its policy decisions that correspond to changing several right-hand sides simultaneously in a variety of ways. Although the spreadsheet can be used to see the effect of any combination of simultaneous changes, it can take an exorbitant amount of time to systematically investigate a large number of simultaneous changes in right-hand sides in this way. Fortunately, Solver’s sensitivity report provides valuable information for guiding such an investigation. In particular, there is a 100 percent rule (analogous to the one presented in Section 5.4) that uses this information to conduct this kind of investigation. Recall that the 100 percent rule described in Section 5.4 is used to investigate simultaneous changes in objective function coefficients. The new 100 percent rule presented next investigates simultaneous changes in right-hand sides in a similar way. The data needed to apply the new 100 percent rule for the Wyndor problem are given by the last three columns in the bottom part of the sensitivity report in Figure 5.21. Keep in mind that we can safely use the entire allowable decrease or increase from the current value of a right-hand side only if none of the other right-hand sides are changed at all. With simultaneous changes in the right-hand sides, we focus for each change on the percentage of the allowable decrease or increase that is being used. As detailed next, the 100 percent rule basically says that we can safely make the simultaneous changes only if the sum of these percentages does not exceed 100 percent. The 100 Percent Rule for Simultaneous Changes in Right-Hand Sides: The shadow prices remain valid for predicting the effect of simultaneously changing the right-hand sides of some of the functional constraints as long as the changes are not too large. To check whether the changes are small enough, calculate for each change the percentage of the allowable change (decrease or increase) for that right-hand side to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid. (If the sum does exceed 100 percent, then we cannot be sure.) To illustrate this rule, consider again the simultaneous changes (shifting one hour of production time per week from Plant 3 to Plant 2) that led to Figure 5.23. The calculations for the 100 percent rule in this case are RHS 2: 12 → 13 13 − 12 2 Percentage of allowable increase = 100 _ ( 6 ) = 16⁄3% R HS 3: 18 → 17 18 − 17 2 Percentage of allowable decrease = 100 _ ( 6 ) = 16⁄3% ¯ Sum = 33 1⁄3% 5.7 Robust Optimization 179 Since the sum of 331⁄3percent is less than 100 percent, the shadow prices definitely are valid for predicting the effect of these changes, as was illustrated with Figure 5.23. The fact that 331⁄3percent is one-third of 100 percent suggests that the changes can be three times as large as above without invalidating the shadow prices. To check this, let us apply the 100 percent rule with these larger changes. RHS 2: 12 → 15 15 − 12 Percentage of allowable increase = _ % = 50% ( 6 ) HS 3: 18 → 15 R 18 − 15 Percentage of allowable decrease = _ % = 50% ( 6 ) Sum = ¯ 100% Because the sum does not exceed 100 percent, the shadow prices are still valid, but these are the largest changes in the right-hand sides that can provide this guarantee. In fact, Figure 5.25 demonstrated that the shadow prices become invalid for larger changes. Review Questions 5.7 1. Why might it be of interest to investigate the effect of making simultaneous changes in the functional constraints? 2. How can the spreadsheet be used to investigate simultaneous changes in the functional constraints? 3. What are the capabilities of a parameter analysis report for investigating simultaneous changes in the functional constraints? 4. Why might a manager be interested in considering simultaneous changes in right-hand sides? 5. What is the 100 percent rule for simultaneous changes in right-hand sides? 6. What are the data needed to apply the 100 percent rule for simultaneous changes in righthand sides? 7. What is guaranteed if the sum of the percentages of allowable changes in the right-hand sides does not exceed 100 percent? 8. What is the conclusion if the sum of the percentages of allowable changes in the right-hand sides does exceed 100 percent? ROBUST OPTIMIZATION As described in the preceding sections, what-if analysis provides an important way of dealing with uncertainty about the true values of the parameters in a linear programming model. One main purpose of what-if analysis is to identify each of the sensitive parameters, namely, each of the parameters where even a small change in its value can change the optimal solution. This is valuable information since these are the parameters that need to be estimated with special care to minimize the risk of obtaining an erroneous optimal solution. Unfortunately, it often is not possible to estimate the sensitive parameters with as much accuracy as desired. The true values of these parameters may not become known until considerably later when the optimal solution (according to the model) is actually implemented. Therefore, even after estimating the sensitive parameters as carefully as possible, significant estimation errors can occur for these parameters along with even larger estimation errors for the other parameters. This can lead to unfortunate consequences. Perhaps the optimal solution (according to the model) will not be optimal after all. In fact, it may not even be feasible. The seriousness of these unfortunate consequences depends somewhat on whether there is any latitude in the functional constraints in the model. It is useful to make the following distinction between these constraints. A soft constraint is a constraint that can be violated a little bit without very serious complications. By contrast, a hard constraint is a constraint that must be satisfied. Robust optimization is especially designed for dealing with problems with hard constraints. 180 Chapter Five What-If Analysis for Linear Programming The goal of robust optimization is to find a solution for the model that is guaranteed to remain feasible and near optimal for all plausible combinations of the actual values for the parameters. Such a solution is called a robust solution. This is a daunting goal, but an elaborate theory of robust optimization now has been developed. Much of this theory is beyond the scope of this book, but we will introduce the basic concept by considering the following straightforward case of independent parameters. Robust Optimization with Independent Parameters This case makes four basic assumptions, where assumption 3 defines independent parameters. 1. Each parameter has a range of uncertainty surrounding its estimated value. 2. This parameter can take any value between the minimum and maximum specified by this range of uncertainty. 3. This value is uninfluenced by the values taken on by the other parameters. 4. All the functional constraints are in either ≤ or ≥ form. Generally, in robust optimization, each uncertain parameter should be assigned its most conservative value within the range of uncertainty. To guarantee that the solution will remain feasible regardless of the values taken on by these parameters within their ranges of uncertainty, we just need to assign the most conservative value to each uncertain parameter and then find the optimal solution for the revised problem. Example We now will illustrate the robust optimization approach for this case of independent parameters by addressing the fifth what-if question posed by Wyndor management. Question 5: What happens if the production rates of doors and windows at plant 3 are uncertain? In particular, because of the brand new production processes used to assemble the windows and doors at plant 3, management is unsure exactly how many hours will be required to produce each window and each door. The initial estimates (as given in Table 2.1) of 3 hours per door and 2 hours per window could be off by as much as half an hour each. In this case, we are dealing with the following two uncertain parameters: HD3 = production hours required per door in plant 3 HW3 = production hours required per window in plant 3 Because the production processes are different for doors and windows, the value of each of these parameters is uninfluenced by the value of the other one, so these are independent parameters. Since the initial estimates of HD3 = 3 and HW3 = 2 could each be off by as much as half an hour, their ranges of uncertainty are Range of uncertainty for HD3 = 2.5–3.5 hours Range of uncertainty for HW3 = 1.5–2.5 hours By setting all of the uncertain parameters at their most conservative values and running Solver again for the revised problem, a robust solution has been found that is guaranteed to be feasible for all plausible combinations of actual values for the uncertain parameters. The actual values of these parameters will not become known until some time after production has begun, but a decision is needed now regarding the initial production schedule for the doors and windows in the three plants. Recall that the model for the Wyndor problem includes a constraint that the total number of hours devoted to doors and windows in plant 3 cannot exceed 18 hours per week. This is a hard constraint (it must be satisfied), as are the corresponding constraints for plants 1 and 2. Therefore, management would like to choose a production rate for windows and doors that is guaranteed to be remain feasible and also should be at least nearly optimal. Applying the procedure for robust optimization with independent parameters, each uncertain parameter is assigned the most conservative value in its range of uncertainty. It is most conservative to assume each door requires the maximum value in its range of uncertainty, or 3.5 hours. Similarly, it is most conservative to assume each window requires 2.5 hours. 5.7 Robust Optimization 181 FIGURE 5.26 A Applying robust optimization to the Wyndor problem, each uncertain parameter has been set to its most conservative value in its range of uncertainty: HD3 = 3.5 and HW3 = 2.5. Running Solver then yields this robust solution. 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0.857 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3.5 2.5 18 ≤ 18 Doors Windows Total Profit 0.857 6 $3,257 10 11 12 Units Produced Making this change in the spreadsheet model and running Solver yields the spreadsheet shown in Figure 5.26. Producing 0.857 doors and 6 windows per week yields a profit of $3,257 and is guaranteed to be feasible if the hours required per door and window fall anywhere within their ranges of uncertainty. The General Procedure for Robust Optimization with Independent Parameters The above example considers a case whether the only uncertain parameters were the coefficients in a single functional constraint. However, robust optimization with independent parameters can also be implemented when there is a range of uncertainty for various other parameters in the model, including objective coefficients and the right-hand-side of the constraints. Simply choose the most conservative value for each parameter as follows: • For each functional constraint in ≤ form, use the maximum value for each uncertain ­coefficient on the left of the ≤ and the minimum value for an uncertain right-hand-side. • For each functional constraint in ≥ form, use the minimum value for each uncertain ­coefficient on the left of the ≥ and the maximum value for an uncertain right-hand-side. • For an objective function in maximization form, use the minimum value of each uncertain coefficient. • For an objective function in minimization form, use the maximum value for each uncertain coefficient. Then run Solver again for the revised model. The resulting solution is guaranteed to be feasible so long as all parameters fall somewhere within their ranges of uncertainty. Review Questions 1. 2. 3. 4. 5. What is meant by a soft constraint? What is meant by a hard constraint? What is the goal of robust optimization? What are the four basic assumptions of robust optimization with independent parameters? In the general procedure for robust optimization with independent parameters, how should the value of an uncertain coefficient on the left-hand-side of a functional constraint be chosen from within its range of uncertainty? 6. In the general procedure for robust optimization with independent parameters, how should the value of an uncertain coefficient in the objective function be chosen from within its range of uncertainty? 7. In the general procedure for robust optimization with independent parameters, how should the value of an uncertain right-hand-side of a functional constraint be chosen from within its range of uncertainty? 182 Chapter Five What-If Analysis for Linear Programming 5.8 CHANCE CONSTRAINTS WITH ANALYTIC SOLVER Two shortcomings of robust optimization are that (1) it can be extremely conservative, and (2) it can be difficult to accurately identify an upper and lower bound for an uncertain parameter. Chance constraints are useful when dealing with soft constraints. The parameters of a linear programming model often remain uncertain until the actual values of these parameters can be observed at some later time when the adopted solution is implemented for the first time. The preceding section described how robust optimization deals with this uncertainty by revising the values of the parameters in the model to ensure that the resulting solution will be feasible when it finally is implemented. When dealing with independent parameters, this involves identifying an upper and lower bound on the possible value of each uncertain parameter. The estimated value of the parameter then is replaced by whichever of these two bounds is the more conservative one. This is a useful approach when dealing with hard constraints, i.e., those constraints that must be satisfied. However, it does have certain shortcomings. One is that it can be extremely conservative in tightening the model far more than is realistically necessary. This is especially true when dealing with large models with hundreds or thousands (perhaps even millions) of parameters. Another is that it might not be possible to accurately identify an upper and lower bound for an uncertain parameter. In fact, it might not even have an upper and lower bound. This is the case, for example, when the underlying probability distribution for a parameter is a normal distribution, which has long tails with no bounds. Chance constraints are useful when dealing with soft constraints, i.e., those constraints that can be violated a little bit without very serious complications. Therefore, rather than requiring that a soft constraint must be satisfied, it would make sense to require only that it probably will be satisfied but there is a small chance that it will be violated a little bit. This is what a chance constraint does. A chance constraint only requires that the original constraint will be satisfied with some very high probability while leaving a small chance that it will be violated a little bit. The Analytic Solver includes the ability to define uncertain parameters. It also can include chance constraints. To implement chance constraints in Analytic Solver, each of the uncertain parameters must first be specified as following a particular probability distribution. This concept will be covered in much more detail in Chapter 13 (Computer Simulation with Analytic Solver). For the purpose of demonstrating chance constraints now, a simple formula for specifying that a parameter should follow the uniform distribution (and later the normal distribution) will be shown. Chapter 13 will consider various more advanced probability distributions and other methods of defining them using the menus on the Analytic Solver ribbon. Chance Constraints with the Uniform Distribution We now will illustrate chance constraints by again addressing the fifth what-if question posed by Wyndor management in the preceding section. Question 5: What happens if the production rates of doors and windows at plant 3 are uncertain? As discussed in the preceding section, the introduction of brand new production processes for assembling the windows and doors in plant 3 results in uncertainty about how many hours will be required to assemble each window and each door there. The range of uncertainty for the hours required per door is 2.5–3.5, while the range of uncertainty for the hours required per window is 1.5–2.5. If we assume that any value within these ranges is equally likely, this would mean that each of these parameters follows the uniform distribution. Analytic Solver includes a function called PsiUniform(Min, Max) to specify that a value should follow the uniform distribution over the range from Min to Max. As shown in ­Figure 5.27, =PsiUniform(2.5, 3.5) is entered in cell C9 and =PsiUniform(1.5, 2.5) is entered in cell D9. Analytic Solver then generates a random value from the specified uniform distribution (shown as 2.737 and 2.317 in cells C9 and D9, respectively). In the original Wyndor problem, it was optimal to produce 2 doors per week and 6 windows per week. However, with the uncertainty in the production rates at plant 3, this solution could lead to using too many hours at plant 3. In fact, this happens for the particular random values that Analytic Solver has generated in cells C9 and D9 of Figure 5.27. At these production 5.8 Chance Constraints with Analytic Solver 183 FIGURE 5.27 A spreadsheet model for applying chance constraints to the Wyndor problem before finding a final solution. The hours used per door and per window produced have been specified to follow the uniform distribution in cells C9 and D9, respectively. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 2 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 2.737 2.317 19.375 ≤ 18 Doors Windows Total Profit 2 6 $3,600 10 11 12 Units Produced 9 C D =PsiUniform(2.5,3.5) =PsiUniform(1.5,2.5) rates, 19.375 hours would be used at plant 3 while only 18 hours are available. There still is a little more work that needs to be done before finding a final solution. Wyndor management still needs to define a chance constraint for plant 3. Then Analytic Solver will be used to solve the model to find a solution such that the original constraint would usually be satisfied. Robust optimization was used in the preceding section to find a solution that guarantees a feasible solution, with no more than 18 hours used per week in plant 3. However, management now realizes there may be some flexibility with this constraint. For example, there may actually be some flexibility with the number of hours available for doors and windows in plant 3. If more than 18 hours were required in some week, overtime could be used to meet the requirements. However, overtime is expensive and management would like to avoid this most of the time. Therefore, in contrast to what was done with robust optimization in the preceding section, management will no longer absolutely require that this constraint be met. Instead, they have decided that they would like to require that this constraint be satisfied at least 95 percent of the time. This can be done using a chance constraint. To implement this chance constraint in Analytic Solver, choose cell E9 (hours used in plant 3 in Figure 5.27), and then select <= from Constraints > Chance Constraints on the Analytic Solver ribbon. This brings up the dialog box shown in Figure 5.28. Cell E9 is already chosen as the left-hand side of the constraint. Choose cell G9 for the right-hand side of the constraint. Finally, under Chance, choose the probability that the constraint should be satisfied. The default value of 0.95 is entered in Figure 5.28. Click OK to accept this chance constraint. An important option for chance constraints in Analytic Solver is found under the Platform tab in the Model pane. Auto Adjust Chance Constraints, as shown in Figure 5.29, can be set to either True or False (and is False by default). Set at False, the solution will often be very conservative in the sense that it usually will give a probability that actually is considerably FIGURE 5.28 This Add Constraint dialog box specifies a chance constraint that E9 is required to be less than or equal to G9 with probability 0.95. 184 Chapter Five What-If Analysis for Linear Programming FIGURE 5.29 The Platform tab of Analytic Solver’s Model pane that shows the Auto Adjust Chance Constraints option set to True. Setting the Auto Adjust Chance Constraints option to True results in a better solution while still meeting the requirements of the chance constraints. Generally, robust optimization yields a more conservative solution than when using chance constraints with uniform distributions. higher than the one specified in the chance box shown in Figure 5.28. Setting this option to True causes Solver to automatically re-optimize the initial conservative solution to find a better solution that still meets the requirements of each chance constraint with the probability specified in the chance box. Therefore, this option generally should be set to True to find better (albeit less conservative) solutions. With Auto Adjust Chance Constraints set to true, the model is then solved by clicking on the Optimize button on the Analytic Solver ribbon. Analytic Solver will then find a solution that maximizes TotalProfit (G12) while meeting all the hard constraints (E7:E8 ≤ G7:G8) and yields at least a 95% chance of satisfying the original constraint (E9 ≤G9). This solution is shown in Figure 5.30. It is interesting to compare the solution found by robust optimization and shown in ­Figure 5.26 with the solution shown in Figure 5.30 when a chance constraint has been used. With robust optimization (requiring that no more than 18 hours ever will be used in any week), 0.857 doors and 6 windows are produced per week, yielding a profit of $3,257. When using a chance constraint (requiring just a 95% chance that no more than 18 hours are used in plant 3 in any given week), 1.103 doors and 6 windows should be produced per week, yielding a higher profit of $3,331. This comparison between the two approaches is typical in other cases as well. In general, robust optimization yields a very conservative solution while chance constraints with uniform distributions allow for a better solution if there is some flexibility in the original constraints. Chance Constraints with the Normal Distribution The robust optimization procedure discussed in the preceding section requires that each uncertain parameter have a specified range of uncertainty (a minimum and a maximum). While the FIGURE 5.30 The spreadsheet obtained after solving the Wyndor problem with Analytic Solver to maximize the TotalProfit (G12) while meeting the hard constraints (E7:E8 ≤ G7:G8) and having at least a 95% chance of satisfying the original constraint (E9 ≤ G9). A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 1.1027 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 3.108 2.193 16.584 ≤ 18 Doors Windows Total Profit 1.103 6 $3,331 10 11 12 Units Produced 5.8 Chance Constraints with Analytic Solver 185 A more conservative solution is necessary to satisfy a chance constraint when the variability is increased in the corresponding parameter. Review Questions uniform distribution for an uncertain parameter meets the requirement of having a range of uncertainty, many other distributions do not. For example, the normal distribution has long tails with no minimum or maximum. Therefore, the procedure for robust optimization discussed in the preceding section cannot be used. However, chance constraints can still be used. Suppose management is now more uncertain about the hours used per door and window at plant 3. The uniform distribution assumed a maximum deviation of 0.5 hours from the original estimate, so the range for this distribution is 2.5–3.5 hours for doors and 1.5–2.5 hours for windows. Management now feels that these parameters should follow the normal distribution, with a standard deviation of 0.5 hours. Given the long tails of the normal distribution, this leads to the possibility that the hours used could deviate well more than 0.5 hours from the original estimates. Analytic Solver includes a function called PsiNormal(Mean, Standard Deviation) to specify the normal distribution. Figure 5.31, shows the model revised to use the normal distribution, with =PsiNormal(3, 0.5) entered in cell C9 and =PsiNormal(2, 0.5) entered in cell D9. Analytic Solver has generated a random value from the specified normal distribution (shown as 2.543 and 2.256 in cells C9 and D9, respectively). The chance constraint is defined in the same way as with the uniform distribution, with 0.95 again entered into the chance box. After clicking the Optimize button on the Analytic Solver ribbon, Figure 5.31 shows the new optimal solution. Now 0.298 doors and 6 windows should be produced per week, leading to a profit of $3,089. This leads to a 95% chance of satisfying the original constraint for plant 3. It is interesting to compare the solution in Figure 5.30 (which assumes uniform distributions with a maximum deviation of 0.5 hours) to the solution in Figure 5.31 (which assumes normal distributions with a standard deviation of 0.5 hours). With normal distributions, just 0.298 doors are produced per week instead of 1.103 when using uniform distributions, leading to a smaller profit of $3,089 as compared to $3,331. This more conservative solution is necessary to maintain the 95% probability of satisfying the original constraint for plant 3 given the higher variability in the production rates with this particular normal distribution. (A normal distribution with a considerably smaller standard deviation would lead to a considerably less conservative solution.) 1. Are chance constraints more useful for hard constraints or soft constraints? 2. What Analytic Solver function is used to specify a value that follows the uniform distribution? 3. Which approach will lead to a more conservative solution—robust optimization with specified ranges of uncertainty or using chance constraints with uniform distributions over these same ranges of uncertainty? 4. What Analytic Solver function is used to specify a value that follows the normal distribution? FIGURE 5.31 The spreadsheet obtained after solving the Wyndor problem with a chance constraint for plant 3 when the hours used per door and per window produced have been specified to follow the normal distribution. A 1 B C D E F G Wyndor Glass Co. Product-Mix Problem 2 3 4 Unit Profit Doors Windows $300 $500 5 6 Hours Used per Unit Produced Hours Hours Used Available 7 Plant 1 1 0 0.2982 ≤ 4 8 Plant 2 0 2 12 ≤ 12 9 Plant 3 2.543 2.256 14.295 ≤ 18 Doors Windows Total Profit 0.298 6 $3,089 10 11 12 Units Produced C 9 =PsiNormal(3, 0.5) D =PsiNormal(2, 0.5) 186 Chapter Five What-If Analysis for Linear Programming 5.9 Summary What-if analysis is analysis done after finding an optimal solution for the original version of the basic model. This analysis provides important insights to help guide managerial decision making. This chapter describes how this is done when the basic model is a linear programming model. The spreadsheet for the model, the parameter analysis report available with Analytic Solver, and the sensitivity report generated by Solver all play a central role in this process. The coefficients in the objective function typically represent quantities that can only be roughly estimated when the model is formulated. Will the optimal solution obtained from the model be the correct one if the true value of one of these coefficients is significantly different from the estimate used in the model? The spreadsheet can be used to quickly check specific changes in the coefficient. The parameter analysis report enables the systematic investigation of many trial values for this coefficient. For a broader investigation, the allowable range for each coefficient identifies the interval within which the true value must lie in order for this solution to still be the correct optimal solution. These ranges are easily calculated from the data in the sensitivity report provided by Solver. What happens if there are significant inaccuracies in the estimates of two or more coefficients in the objective function? Specific simultaneous changes can be checked with the spreadsheet. A two-way parameter analysis report can systematically investigate various simultaneous changes in two coefficients. To go further, the 100 percent rule for simultaneous changes in objective function coefficients provides a convenient way of checking whole ranges of simultaneous changes, again by using the data in Solver’s sensitivity report. What-if analysis usually extends to considering the effect of changes in the functional constraints as well. Occasionally, changes in the coefficients of these constraints will be considered because of the uncertainty in their original estimates. More frequently, the changes considered will be in the right-hand sides of the constraints. The right-hand sides frequently represent managerial policy decisions. In such cases, shadow prices provide valuable guidance to management about the potential effects of altering these policy decisions. The shadow price for each constraint is easily found by using the spreadsheet, a parameter analysis report, or the sensitivity report. Shadow price analysis can be validly applied to investigate possible changes in right-hand sides as long as these changes are not too large. The allowable range for each right-hand side indicates just how far it can be changed, assuming no other changes are made. If other right-hand sides are changed as well, the 100 percent rule for simultaneous changes in right-hand sides enables checking whether the changes definitely are not too large. Solver’s sensitivity report provides the key information needed to find each allowable range or to apply this 100 percent rule. Both the spreadsheet and the parameter analysis report also can sometimes be used to help investigate these simultaneous changes. In some cases, there may be so much uncertainty about what the true values of the model parameters will turn out to be, management may want to be sure that the proposed solution will at least turn out to be feasible. This is particularly important if it would be completely unacceptable to violate any of the constraints even a little bit. Robust optimization provides a way of identifying a solution that is virtually guaranteed to be feasible as well as nearly optimal regardless of what the true values of the model parameters turn out to be. If the model includes some constraints that actually can be violated a little bit without serious complications, chance constraints can be used to represent these constraints. A chance constraint requires that the original constraint probably will be satisfied but allows a small probability for violating the original constraint by a small amount. Glossary allowable range for an objective function coefficient The range of values for a particular coefficient in the objective function over which the optimal solution for the original model remains optimal. (Section 5.3), 160 allowable range for the right-hand side The range of values for the right-hand side of a functional constraint over which this constraint’s shadow price remains valid. (Section 5.5), 172 chance constraint A modification of a constraint which only requires that the original constraint will be satisfied with some very high probability while leaving a small chance that it will violated a little bit. (Section 5.8), 182 hard constraint A constraint that must be satisfied. (Section 5.7) 179 independent parameters Parameters whose values are uninfluenced by the values taken on by the other parameters. (Section 5.7), 180 parameter cell A data cell containing a parameter that will be systematically varied when generating a parameter analysis report. (Section 5.3), 158 parameters of the model The parameters of a linear programming model are the constants (coefficients or right-hand sides) in the functional constraints and the objective function. (Section 5.1), 152 Chapter 5 Solved Problem 187 range of uncertainty The range between the minimum and maximum possible value for a parameter. (Section 5.7), 180 robust optimization An optimization procedure that finds a solution that is guaranteed to remain feasible and near optimal for all plausible combinations of the actual values for the parameters. (Section 5.7) 180 robust solution A solution found by using robust optimization. (Section 5.7) 180 sensitive parameter A parameter is considered sensitive if even a small change in its value can change the optimal solution. (Section 5.1), 152 sensitivity analysis The part of what-if analysis that focuses on individual parameters of the model. It involves checking how sensitive the optimal solution is to the value of each parameter. (Section 5.1), 153 shadow price The shadow price for a functional constraint is the rate at which the optimal value of the objective function can be increased by increasing the right-hand side of the constraint by a small amount. (Section 5.5), 172 soft constraint A constraint that can be violated a little bit without very serious complications. (Section 5.7), 179 what-if analysis Analysis that addresses questions about what would happen to the optimal solution if different assumptions were made about future conditions. (Chapter introduction), 151 Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Excel Add-in: Excel Files: Analytic Solver Wyndor Example Supplement to This Chapter: Profit & Gambit Example Reduced Costs Solved Problem The solution is available at www.mhhe.com/Hillier6e. 5.S1. Sensitivity Analysis at Stickley Furniture Stickley Furniture is a manufacturer of fine hand-crafted furniture. During the next production period (the eight-hour shift tomorrow), management is considering producing dining room tables, dining room chairs, and/or bookcases. The time required for each item to go through the two stages of production (assembly and finishing), the amount of wood required (fine cherry wood), and the corresponding unit profits are given in the following table, along with the amount of each resource available in the upcoming production period. Tables Chairs Bookcases Available Assembly (minutes) 80 40 50 8,100 Finishing) (minutes) 30 20 30 4,500 Wood (pounds) 80 10 50 9,000 $360 $125 $300 Unit profit a. Suppose the profit per table increases by $100. Will this change the optimal production quantities? What can be said about the change in total profit? b. Suppose the profit per chair increases by $100. Will this change the optimal production quantities? What can be said about the change in total profit? c. Suppose the profit per table increases by $90 and the profit per bookcase decreases by $50. Will this change the optimal production quantities? What can be said about the change in total profit? d. Suppose a worker in the assembly department calls in sick, so eight fewer hours will be available tomorrow in the assembly department. How much would this affect total profit? Would it change the optimal production quantities? e. Explain why the shadow price for the wood constraint is zero. After formulating a linear programming model to determine the production levels that would maximize profit, the solved model and the corresponding sensitivity report are shown below. f. A new worker has been hired who is trained to do both assembly and finishing. She will split her time between the two areas, so there now are four additional hours available tomorrow in both assembly and finishing. How much would this affect total profit? Would this change the optimal production quantities? g. Based on the sensitivity report given below, is it wise to have the new worker in part f split her time equally between assembly and finishing, or would some other plan be better? 188 Chapter Five What-If Analysis for Linear Programming B 3 4 Unit Profit C D E Tables Chairs Bookcases F $360 $125 $300 G H 5 Resources Required per Unit 6 Used Available 7 Assembly (minutes) 80 40 50 8,100 8,100 8 Finishing (minutes) 30 20 30 4,500 4,500 9 Woods (pounds) 80 10 50 8,100 9,000 Tables Chairs Bookcases Total Profit 20 0 130 $46,200 Final Value Reduced Cost 10 11 Production 12 Variable Cells Cell Name $C$12 Production Tables 20 $D$12 Production Chairs 0 $E$12 Production Bookcases Objective Coefficient 0 360 −88.333 130 Allowable Increase 0 Allowable Decrease 120 60 125 88.333 300 60 1E + 30 75 Constraints Final Value Shadow Price Constraint R. H. Side Cell Name $F$7 Assembly (minutes) Used 8,100 2 8,100 $F$8 Finishing (minutes) Used 4,500 6.67 $F$9 Wood (pounds) Used 8,100 0 h. Use Analytic Solver and a parameter analysis report to determine how the optimal production quantities and total profit will change depending on how the new worker in part f allocates her time between assembly and finishing. Allowable Increase Allowable Decrease 900 600 4,500 360 1,462.5 9,000 1E + 30 900 In particular, assume 0, 1, 2, . . . , or 8 hours are added to assembly, with a corresponding 8, 7, 6, . . . , or 0 hours added to finishing. (The original spreadsheet is contained in www.mhhe.com/Hillier6e.) Problems We have inserted the symbol AS to the left of each problem (or its parts) where Analytic Solver is required. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 5.1.* One of the products of the G. A. Tanner Company is a special kind of toy that provides an estimated unit profit of $3. Because of a large demand for this toy, management would like to increase its production rate from the current level of 1,000 per day. However, a limited supply of two subassemblies (A and B) from vendors makes this difficult. Each toy requires two subassemblies of type A, but the vendor providing these subassemblies would only be able to increase its supply rate from the current 2,000 per day to a maximum of 3,000 per day. Each toy requires only one subassembly of type B, but the vendor providing these subassemblies would be unable to increase its supply rate above the current level of 1,000 per day. Because no other vendors currently are available to provide these subassemblies, management is considering initiating a new production process internally that would simultaneously produce an equal number of subassemblies of the two types to supplement the supply from the two vendors. It is estimated that the company’s cost for producing one subassembly of each type would be $2.50 more than the cost of purchasing these subassemblies from the two vendors. Management wants to determine both the production rate of the toy and the production rate of each pair of subassemblies (one A and one B) that would maximize the total profit. Viewing this problem as a resource-allocation problem, one of the company’s managers has organized its data as follows: Resource Usage per Unit of Each Activity Resource Produce Toys Produce Subassemblies Amount of Resource Available Subassembly A 2 −1 3,000 Subassembly B 1 −1 1,000 Unit profit $3 −$2.50 Chapter 5 Problems 189 a. Formulate and solve a spreadsheet model for this problem. b. Since the stated unit profits for the two activities are only estimates, management wants to know how much each of these estimates can be off before the optimal solution would change. Begin exploring this question for the first activity (producing toys) by using the spreadsheet and Solver to manually generate a table that gives the optimal solution and total profit as the unit profit for this activity increases in 50¢ increments from $2.00 to $4.00. What conclusion can be drawn about how much the estimate of this unit profit can differ in each direction from its original value of $3.00 before the optimal solution would change? c. Repeat part b for the second activity (producing subassemblies) by generating a table as the unit profit for this activity increases in 50¢ increments from –$3.50 to –$1.50 (with the unit profit for the first activity fixed at $3). AS d. Use a parameter analysis report to systematically generate all the data requested in parts b and c, except use 25¢ increments instead of 50¢ increments. Use these data to refine your conclusions in parts b and c. e. Use Solver’s sensitivity report to find the allowable range for the unit profit of each activity. AS f. Use a two-way parameter analysis report to systematically generate the total profit as the unit profits of the two activities are changed simultaneously as described in parts b and c. g. Use the information provided by Solver’s sensitivity report to describe how far the unit profits of the two activities can change simultaneously before the optimal solution might change. 5.2. Consider a resource-allocation problem having the following data: Resource Usage per Unit of Each Activity Resource 1 2 Amount of Resource Available 1 1 2 10 1 3 12 $2 $5 2 Unit profit The objective is to determine the number of units of each activity to undertake so as to maximize the total profit. While doing what-if analysis, you learn that the estimates of the unit profits are accurate only to within ± 50 percent. In other words, the ranges of likely values for these unit profits are $1 to $3 for activity 1 and $2.50 to $7.50 for activity 2. a. Formulate a spreadsheet model for this problem based on the original estimates of the unit profits. Then use Solver to find an optimal solution and to generate the sensitivity report. b. Use the spreadsheet and Solver to check whether this optimal solution remains optimal if the unit profit for activity 1 changes from $2 to $1. From $2 to $3. c. Also check whether the optimal solution remains optimal if the unit profit for activity 1 still is $2 but the unit profit for activity 2 changes from $5 to $2.50. From $5 to $7.50. AS d. Use a parameter analysis report to systematically generate the optimal solution and total profit as the unit profit of activity 1 increases in 20 cent increments from $1 to $3 (without changing the unit profit of activity 2). Then do the same as the unit profit of activity 2 increases in 50 cent increments from $2.50 to $7.50 (without changing the unit profit of activity 1). Use these results to estimate the allowable range for the unit profit of each activity. e. Use the sensitivity report to find the allowable range for the unit profit of each activity. Then use these ranges to check your results in parts b-d. AS f. Use a two-way parameter analysis report to systematically generate the optimal total profit as the unit profit of the two activities are changes simultaneously as described in part d. 5.3. Consider the Big M Co. problem presented in Section 3.5, including the spreadsheet in Figure 3.10 showing its formulation and optimal solution. There is some uncertainty about what the unit costs will be for shipping through the various shipping lanes. Therefore, before adopting the optimal solution in Figure 3.10, management wants additional information about the effect of inaccuracies in estimating these unit costs. Use Solver to generate the sensitivity report preparatory to addressing the following questions. a. Which of the unit shipping costs given in Table 3.9 has the smallest margin for error without invalidating the optimal solution given in Figure 3.10? Where should the greatest effort be placed in estimating the unit shipping costs? b. What is the allowable range for each of the unit shipping costs? c. How should the allowable range be interpreted to management? d. If the estimates change for more than one of the unit shipping costs, how can you use the sensitivity report to determine whether the optimal solution might change? 5.4.* Consider the Union Airways problem presented in Section 3.3, including the spreadsheet in Figure 3.5 showing its formulation and optimal solution. Management is about to begin negotiations on a new contract with the union that represents the company’s customer service agents. This might result in some small changes in the daily costs per agent given in Table 3.5 for the various shifts. Several possible changes listed below are being considered separately. In each case, management would like to know whether the change might result in the solution in Figure 3.5 no longer being optimal. Answer this question in parts a to e by using the spreadsheet and Solver directly. If the optimal solution changes, record the new solution. a. The daily cost per agent for shift 2 changes from $160 to $165. 190 Chapter Five What-If Analysis for Linear Programming b. The daily cost per agent for shift 4 changes from $180 to $170. c. The changes in parts a and b both occur. d. The daily cost per agent increases by $4 for shifts 2, 4, and 5, but decreases by $4 for shifts 1 and 3. e. The daily cost per agent increases by 2 percent for each shift. f. Use Solver to generate the sensitivity report for this problem. Suppose that the above changes are being considered later without having the spreadsheet model immediately available on a computer. Show in each case how the sensitivity report can be used to check whether the original optimal solution must still be optimal. AS g. For each of the five shifts in turn, use a parameter analysis report to systematically generate the optimal solution and total cost when the only change is that the daily cost per agent on that shift increases in $3 increments from $15 less than the current cost up to $15 more than the current cost. 5.5. Consider the Think-Big Development Co. problem presented in Section 3.2, including the spreadsheet in Figure 3.3 showing its formulation and optimal solution. In parts a-g, use the spreadsheet and Solver to check whether the optimal solution would change and, if so, what the new optimal solution would be, if the estimates in Table 3.3 of the net present values of the projects were to be changed in each of the following ways. (Consider each part by itself.) a. The net present value of project 1 (a high-rise office building) increases by $200,000. b. The net present value of project 2 (a hotel) increases by $200,000. c. The net present value of project 1 decreases by $5 million. d. The net present value of project 3 (a shopping center) decreases by $200,000. e. All three changes in parts b, c, and d occur simultaneously. f. The net present values of projects 1, 2, and 3 change to $46 million, $69 million, and $49 million, respectively. g. The net present values of projects 1, 2, and 3 change to $54 million, $84 million, and $60 million, respectively. A 1 2 3 4 5 6 7 8 9 10 Unit Profit Molding (minutes) Finishing (minutes) Clay (ounces) Production B Plates $3.10 4 8 5 h. Use Solver to generate the sensitivity report for this problem. For each of the preceding parts, suppose that the change occurs later without having the spreadsheet model immediately available on a computer. Show in each case how the sensitivity report can be used to check whether the original optimal solution must still be optimal. AS i. For each of the three projects in turn, use a parameter analysis report to systematically generate the optimal solution and the total net present value when the only change is that the net present value of that project increases in $1 million increments from $5 million less than the current value up to $5 million more than the current value. 5.6. University Ceramics manufactures plates, mugs, and steins that include the campus name and logo for sale in campus bookstores. The time required for each item to go through the two stages of production (molding and finishing), the material required (clay), and the corresponding unit profits are given in the following table, along with the amount of each resource available for tomorrow’s eight-hour shift. Plates Mugs Steins Available Molding (minutes) 4 6 3 2,400 Finishing (minutes) 8 14 12 7,200 Clay (ounces) 5 4 3 3,000 $3.10 $4.75 $4.00 Unit Profit A linear programming model has been formulated in a spreadsheet to determine the production levels for tomorrow that would maximize profit. The solved spreadsheet model and corresponding sensitivity report are shown below. For each of the following parts, answer the question as specifically and completely as is possible without re-solving the problem with Solver. Note: Each part is independent (i.e., any change made in one part does not apply to any other parts). a. Suppose the profit per plate decreases from $3.10 to $2.80. Will this change the optimal production quantities? What can be said about the change in total profit? b. Suppose the profit per stein increases by $0.30 and the profit per plate decreases by $0.25. Will this change the optimal production quantities? What can be said about the change in total profit? C Mugs $4.75 Resource Required per Unit 6 14 4 Plates 300 Mugs 0 D Steins $4.00 3 12 3 Steins 400 E Used 2,400 7,200 2,700 F G <= <= <= Available 2,400 7,200 3,000 Total Profit $2,530 Chapter 5 Problems 191 Variable Cells Cell Final Value Name $B$10 Production Plates 300 $C$10 Production Mugs 0 $D$10 Production Steins 400 Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease 3.10 2.23 0.37 4.75 0.46 4.00 0.65 1.37 Allowable Increase Allowable Decrease 0 −0.46 0 Constraints Cell Name Final Value Shadow Price $E$5 Molding (minutes) Used 2,400 0.22 2400 200 600 $E$6 Finishing (minutes) Used 7,200 0.28 7200 2400 2400 $E$7 Clay (ounces) Used 2,700 0 3000 1E + 30 c. Suppose a worker in the molding department calls in sick. Now eight fewer hours are available tomorrow in the molding department. How much would this affect total profit? Would it change the optimal production quantities? d. Suppose one of the workers in the molding department is also trained to do finishing. Would it be a good idea to have this worker shift some of her time from the molding department to the finishing department? Indicate the rate at which this would increase or decrease total profit per minute shifted. How many minutes can be shifted before this rate might change? e. The allowable decrease for the mugs’ objective coefficient and for the available clay constraint are both missing from the sensitivity report. What numbers should be there? Explain how you were able to deduce each number. 5.7. Ken and Larry, Inc., supplies its ice cream parlors with three flavors of ice cream: chocolate, vanilla, and banana. Due to extremely hot weather and a high demand for its products, the company has run short of its supply of ingredients: milk, sugar, and cream. Hence, they will not be able to fill all the orders received from their retail outlets, the ice cream parlors. Due to these circumstances, the company has decided to choose the amount of each flavor to produce that will maximize total profit, given the constraints on the supply of the basic ingredients. The chocolate, vanilla, and banana flavors generate, respectively, $1.00, $0.90, and $0.95 of profit per gallon sold. The A 1 2 Unit Profit Constraint R. H. Side company has only 200 gallons of milk, 150 pounds of sugar, and 60 gallons of cream left in its inventory. The linear programming formulation for this problem is shown below in algebraic form. Let C = Gallons of chocolate ice cream produced V = Gallons of vanilla ice cream produced B = Gallons of banana ice cream produced Maximize Profit = 1.00C + 0.90V + 0.95B subject to Milk: 0.45 C + 0.50 V + 0.40 B ≤ 200 gallons Sugar: 0.50 C + 0.40 V + 0.40 B ≤ 150 pounds Cream: 0.10 C + 0.15 V + 0.20 B ≤ 60 gallons and C ≥ 0 V ≥ 0 B ≥ 0 This problem was solved using Solver. The spreadsheet (already solved) and the sensitivity report are shown below. (Note: The numbers in the sensitivity report for the milk constraint are missing on purpose, since you will be asked to fill in these numbers in part f.) For each of the following parts, answer the question as specifically and completely as possible without solving the problem again with Solver. Note: Each part is independent (i.e., any change made to the model in one part does not apply to any other parts). B C D Chocolate Vanilla Banana $1.00 $0.90 $0.95 E F G 3 4 Resource Resources Used per Gallon Produced Used Available 5 Milk 0.45 0.5 0.4 180 200 6 Sugar 0.5 0.4 0.4 150 150 7 Cream 0.1 0.15 0.2 60 60 Chocolate Vanilla Banana Total Profit 0 300 75 $341.25 8 9 10 Gallons Produced 192 Chapter Five What-If Analysis for Linear Programming Variable Cells Cell Final Value Name Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease −0.0375 1 0.0375 1E + 30 $B$10 Gallons Produced Chocolate 0 $C$10 Gallons Produced Vanilla 300 0 0.9 0.05 0.0125 $D$10 Gallons Produced Banana 75 0 0.95 0.0214 0.05 Constraints Cell Name Final Value Shadow Price Constraint R. H. Side Allowable Increase 1.875 150 10 30 60 15 3.75 $E$5 Milk Used $E$6 Sugar Used 150 $E$7 Cream Used 60 1 a. What is the optimal solution and total profit? b. Suppose the profit per gallon of banana changes to $1.00. Will the optimal solution change and what can be said about the effect on total profit? c. Suppose the profit per gallon of banana changes to 92¢. Will the optimal solution change and what can be said about the effect on total profit? d. Suppose the company discovers that three gallons of cream have gone sour and so must be thrown out. Will the optimal solution change and what can be said about the effect on total profit? e. Suppose the company has the opportunity to buy an additional 15 pounds of sugar at a total cost of $15. Should it do so? Explain. f. Fill in all the sensitivity report information for the milk constraint, given just the optimal solution for the problem. Explain how you were able to deduce each number. 5.8. Colonial Furniture produces hand-crafted colonial style furniture. Plans are now being made for the production of rocking chairs, dining room tables, and/or armoires over the next week. These products go through two stages of production (assembly and finishing). The table in the next column gives the time required for each item to go through these two stages, the amount of wood required (fine cherry wood), and the corresponding unit profits, along with the amount of each resource available next week. A 1 2 3 4 5 6 7 8 9 10 11 12 Unit Profit Assembly (minutes) Finishing (minutes) Wood (pounds) Production B Rocking Chair $240 Rocking Chair Dining Room Table Armoire Available Assembly (minutes) 100 180 120 3,600 Finishing (minutes) 60 80 80 2,000 Wood (pounds) 30 180 120 4,000 $240 $720 $600 Unit Profit A linear programming model has been formulated in a spreadsheet to determine the production levels that would maximize profit. The solved spreadsheet model and corresponding sensitivity report are shown below. For each of the following parts, answer the question as specifically and completely as is possible without re-solving the problem with Solver. Note: Each part is independent (i.e., any change made in one problem part does not apply to any other parts). a. Suppose the profit per armoire decreases by $50. Will this change the optimal production quantities? What can be said about the change in total profit? b. Suppose the profit per table decreases by $60 and the profit per armoire increases by $90. Will this change the optimal production quantities? What can be said about the change in total profit? C Dining Room Table $720 Resource Required per Unit 100 180 60 80 30 180 Rocking Chair 0 Allowable Decrease Dining Room Table 10 D E F G <= <= <= Available 3,600 2,000 4,000 Armoire $600 120 80 120 Armoire 15 Used 3,600 2,000 3,600 Total Profit $16,200 Chapter 5 Problems 193 Variable Cells Cell Name Final Value Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease $B$12 Production Chair 0 −230 240 230 1E + 30 $C$12 Production Table 10 0 720 180 120 $D$12 Production Armoire 15 0 600 120 120 Constraints Cell Name Final Value Shadow Price Constraint R. H. Side Allowable Increase Allowable Decrease $E$6 Assembly (minutes) Used 3,600 2.00 3,600 400 600 $E$7 Finishing (minutes) Used 2,000 4.50 2,000 400 400 $E$8 Wood (pounds) Used 3,600 c. Suppose a part-time worker in the assembly department calls in sick for one day, so that now four fewer hours are available that day in the assembly department. How much would this affect total profit? Would it change the optimal production quantities? d. Suppose one of the workers in the assembly department is also trained to do finishing. Would it be a good idea to have this worker shift some of his time from the assembly department to the finishing department? Indicate the rate at which this would increase or decrease total profit per minute shifted. How many minutes can be shifted before this rate might change? e. The shadow price and allowable range for the wood constraint are missing from the sensitivity report. What numbers should be there? Explain how you were able to deduce each number. 5.9. David, LaDeana, and Lydia are the sole partners and workers in a company that produces fine clocks. David and LaDeana are each available to work a maximum of 40 hours per week at the company, while Lydia is available to work a maximum of 20 hours per week. The company makes two different types of clocks: a grandfather clock and a wall clock. To make a clock, David (a mechanical engineer) assembles the inside mechanical parts of the clock while LaDeana (a woodworker) produces the hand-carved wood casings. Lydia is responsible for taking orders and shipping the clocks. The amount of time required for each of these tasks is shown next. Time Required Task Grandfather Clock Wall Clock Assemble clock mechanism 6 hours 4 hours Carve wood casing 8 hours 4 hours Shipping 3 hours 3 hours Each grandfather clock built and shipped yields a profit of $300, while each wall clock yields a profit of $200. 4,000 The three partners now want to determine how many clocks of each type should be produced per week to maximize the total profit. a. Formulate a linear programming model in algebraic form for this problem. b. Use graphical analysis to solve the model. Then check if the optimal solution would change if the unit profit for grandfather clocks were changed from $300 to $375 (with no other changes in the model). Then check if the optimal solution would change if, in addition to this change in the unit profit for the grandfather clocks, the estimated unit profit for wall clocks also changed from $200 to $175. c. Formulate and solve the original version of this model on a spreadsheet. d. Use Solver to check the effect of the changes specified in part b. AS e. Use a parameter analysis report to systematically generate the optimal solution and total profit as the unit profit for grandfather clocks is increased in $20 increments from $150 to $450 (with no change in the unit profit for wall clocks). Then do the same as the unit profit for wall clocks is increased in $20 increments from $50 to $350 (with no change in the unit profit for grandfather clocks). Use this information to estimate the allowable range for the unit profit of each type of clock. AS f. Use a two-way parameter analysis report to systematically generate the optimal total profit as the unit profits for the two types of clocks are changed simultaneously as specified in part e, except use $50 increments instead of $20 increments. g. For each of the three partners in turn, use Solver to determine the effect on the optimal solution and the total profit if that partner alone were to increase his or her maximum number of work hours available per week by 5 hours. AS h. Use a parameter analysis report to systematically generate the optimal solution and the total profit 194 Chapter Five What-If Analysis for Linear Programming when the only change is that David’s maximum number of hours available to work per week changes to each of the following values: 35, 37, 39, 41, 43, 45. Then do the same when the only change is that LaDeana’s maximum number of hours available to work per week changes in the same way. Then do the same when the only change is that Lydia’s maximum number of hours available to work per week changes to each of the following values: 15, 17, 19, 21, 23, 25. i. Generate a sensitivity report and use it to determine the allowable range for the unit profit for each type of clock and the allowable range for the maximum number of hours each partner is available to work per week. j. To increase the total profit, the three partners have agreed that one of them will slightly increase the maximum number of hours available to work per week. The choice of which one will be based on which one would increase the total profit the most. Use the sensitivity report to make this choice. (Assume no change in the original estimates of the unit profits.) k. Explain why one of the shadow prices is equal to zero. l. Can the shadow prices in the sensitivity report be validly used to determine the effect if Lydia were to change her maximum number of hours available to work per week from 20 to 25? If so, what would be the increase in the total profit? m. Repeat part l if, in addition to the change for Lydia, David also were to change his maximum number of hours available to work per week from 40 to 35. n. Use graphical analysis to verify your answer in part m. 5.10.* Reconsider Problem 5.1. After further negotiations with each vendor, management of the G. A. Tanner Company has learned that either of them would be willing to consider increasing their supply of their respective subassemblies over the previously stated maxima (3,000 subassemblies of type A per day and 1,000 of type B per day) if the company would pay a small premium over the regular price for the extra subassemblies. The size of the premium for each type of subassembly remains to be negotiated. The demand for the toy being produced is sufficiently high that 2,500 per day could be sold if the supply of subassemblies could be increased enough to support this production rate. Assume that the original estimates of unit profits given in Problem 5.1 are accurate. a. Formulate and solve a spreadsheet model for this problem with the original maximum supply levels and the additional constraint that no more than 2,500 toys should be produced per day. b. Without considering the premium, use the spreadsheet and Solver to determine the shadow price for the subassembly A constraint by solving the model again after increasing the maximum supply by one. Use this shadow price to determine the maximum premium that the company should be willing to pay for each subassembly of this type. c. Repeat part b for the subassembly B constraint. d. Estimate how much the maximum supply of subassemblies of type A could be increased before the shadow price (and the corresponding premium) found in part b would no longer be valid by using a parameter analysis report to generate the optimal solution and total profit (excluding the premium) as the maximum supply increases in increments of 100 from 3,000 to 4,000. AS e. Repeat part d for subassemblies of type B by using a parameter analysis report as the maximum supply increases in increments of 100 from 1,000 to 2,000. f. Use Solver’s sensitivity report to determine the shadow price for each of the subassembly constraints and the allowable range for the right-hand side of each of these constraints. 5.11. Reconsider the model given in Problem 5.2. While doing what-if analysis, you learn that the estimates of the righthand sides of the two functional constraints are accurate only to within ± 50 percent. In other words, the ranges of likely values for these parameters are 5 to 15 for the first right-hand side and 6 to 18 for the second right-hand side. a. After solving the original spreadsheet model, determine the shadow price for the first functional constraint by increasing its right-hand side by one and solving again. AS b. Use a parameter analysis report to generate the optimal solution and total profit as the right-hand side of the first functional constraint is incremented by 1 from 5 to 15. Use this table to estimate the allowable range for this right-hand side, that is, the range over which the shadow price obtained in part a is valid. c. Repeat part a for the second functional constraint. AS d. Repeat part b for the second functional constraint where its right-hand side is incremented by 1 from 6 to 18. e. Use Solver’s sensitivity report to determine the shadow price for each functional constraint and the allowable range for the right-hand side of each of these constraints. 5.12. Consider a resource-allocation problem having the following data: AS Resource Usage per Unit of Each Activity Resource 1 2 Amount of Resource Available 1 1 3 8 2 1 1 4 Unit profit $1 $2 The objective is to determine the number of units of each activity to undertake so as to maximize the total profit. a. Use the graphical method to solve this model. b. Use graphical analysis to determine the shadow price for each of these resources by solving again Chapter 5 Problems 195 after increasing the amount of the resource available by one. c. Use the spreadsheet model and Solver instead to do parts a and b. AS d. For each resource in turn, use a parameter analysis report to systematically generate the optimal solution and the total profit when the only change is that the amount of that resource available increases in increments of 1 from 4 less than the original value up to 6 more than the original value. Use these results to estimate the allowable range for the amount available for each resource. e. Use Solver’s sensitivity report to obtain the shadow prices. Also use this report to find the range for the amount of each resource available over which the corresponding shadow price remains valid. f. Describe why these shadow prices are useful when management has the flexibility to change the amounts of the resources being made available. 5.13. Follow the instructions of Problem 5.12 for a resourceallocation problem that again has the objective of maximizing total profit and that has the following data: Resource Usage per Unit of Each Activity Amount of Resource Available Resource 1 2 1 1 0 4 2 1 3 15 10 3 2 1 Unit profit $3 $2 5.14.* Consider the Super Grain Corp. case study as presented in Section 3.1, including the spreadsheet in Figure 3.1 showing its formulation and optimal solution. Use Solver to generate the sensitivity report. Then use this report to independently address each of the following questions. a. How much could the total expected number of exposures be increased for each additional $1,000 added to the advertising budget? b. Your answer in part a would remain valid for how large of an increase in the advertising budget? c. How much could the total expected number of exposures be increased for each additional $1,000 added to the planning budget? d. Your answer in part c would remain valid for how large of an increase in the planning budget? e. Would your answers in parts a and c definitely remain valid if both the advertising budget and planning budget were increased by $100,000 each? f. If only $100,000 can be added to either the advertising budget or the planning budget, where should it be added to do the most good? g. If $100,000 must be removed from either the advertising budget or the planning budget, from which budget should it be removed to do the least harm? 5.15. Follow the instructions of Problem 5.14 for the continuation of the Super Grain Corp. case study as presented in Section 3.4 including the spreadsheet in Figure 3.7 showing its formulation and optimal solution. 5.16. Consider the Union Airways problem presented in Section 3.3, including the spreadsheet in Figure 3.5 showing its formulation and optimal solution. Management now is considering increasing the level of service provided to customers by increasing one or more of the numbers in the rightmost column of Table 3.5 for the minimum number of agents needed in the various time periods. To guide them in making this decision, they would like to know what impact this change would have on total cost. Use Solver to generate the sensitivity report in preparation for addressing the following questions. a. Which of the numbers in the rightmost column of Table 3.5 can be increased without increasing total cost? In each case, indicate how much it can be increased (if it is the only one being changed) without increasing total cost. b. For each of the other numbers, how much would the total cost increase per increase of 1 in the number? For each answer, indicate how much the number can be increased (if it is the only one being changed) before the answer is no longer valid. c. Do your answers in part b definitely remain valid if all the numbers considered in part b are simultaneously increased by 1? d. Do your answers in part b definitely remain valid if all 10 numbers are simultaneously increased by 1? e. How far can all 10 numbers be simultaneously increased by the same amount before your answers in part b may no longer be valid? 5.17. Reconsider the example illustrating the use of robust optimization that was presented in Section 5.7. Wyndor management now feels that the analysis described in this example was overly conservative for three reasons: (1) it is unlikely that the true values of the parameters, HD3 and HW3, will be as far as half an hour off of the original estimate, (2) it is even more unlikely that both estimates will turn out to simultaneously lean toward the undesirable end of their ranges of uncertainty, and (3) there is a bit of latitude in the constraint to compensate for violating it by a tiny bit. Therefore, Wyndor management has asked its staff (you) to solve the model again while using ranges of uncertainty that are half as wide as those used in Section 5.7. a. Apply the procedure for robust optimization with independent parameters. What is the resulting optimal solution and how much would this increase the total profit per week? b. If Wyndor would need to pay a penalty of $150 per week to the distributor if the production rates fall below these new guaranteed minimum amounts, should Wyndor use these new guarantees? 5.18. Reconsider the example illustrating the use of robust optimization that was presented in Section 5.7. Wyndor management now feels that there is uncertainty in all of the parameters of the problem—the unit profit per door and window (PD and PW), the hours of production time used for each door or window produced across the three plants (HD1, HW2, HD3, and HW3) 196 Chapter Five What-If Analysis for Linear Programming and the three right-hand-sides representing the hours available at each plant (RHS1, RHS2, and RHS3). The original estimates along with the ranges of uncertainty are shown in the table below. Apply the procedure for robust optimization with independent parameters to find the solution that maximizes profit when the solution also is guaranteed to be feasible. Parameter Original Estimate Range of Uncertainty PD $300 $250–$350 PW $500 $400–$600 HD1 1 0.9–1.1 HW2 2 1.6–2.4 HD3 3 2.5–3.5 HW3 2 1.8–2.2 RHS1 4 3.5–4.5 RHS2 12 11–13 RHS3 18 16–20 5.19.* Reconsider Problem 5.12. Now suppose that all of the parameters are uncertain, with ranges of uncertainty as given in the table below. Use the procedure for robust optimization with independent parameters to find the solution that maximizes profit when the solution also is guaranteed to be feasible. Resource Usage per Unit of Each Activity Resource 1 Amount of Resource Available 2 1 0.9–1.1 2.7–3.3 7.2–8.8 2 0.8–1.2 0.7–1.3 3.5–4.5 Unit profit $0.90–$1.10 $1.75–$2.25 5.20. Reconsider Problem 5.13. Now suppose that all of the parameters are uncertain, with ranges of uncertainty as given in the table below. Use the procedure for robust optimization with independent parameters to find the solution that maximizes profit when the solution also is guaranteed to be feasible. Resource Usage per Unit of Each Activity Amount of Resource Available Resource 1 1 0.8–1.2 2 0 3.6–4.4 2 0.9–1.1 2.5–3.5 13.5–16.5 3 1.6–2.4 0.7–1.3 9.5–10.5 Unit profit $2.90–$3.10 $1.80–$2.20 AS 5.21 Reconsider the example illustrating the use of chance constraints that was presented in Section 5.8. Suppose that after a more careful consideration of the hours required per door and window in plant 3, HD3 and HW3, Wyndor management has considerably narrowed the range of what these times might turn out to be. a. HD3 and HW3 are now considered equally likely to be anywhere between 15 minutes below and 15 minutes above the original estimates of 3 hours and 2 hours, respectively. Use a chance constraint and Analytic Solver to find the solution that maximizes total profit while requiring at least a 95% chance of satisfying the original constraint for the third plant. b. Repeat part a, but now require at least a 99% chance of satisfying the constraint at the third plant. c. Now assume HD3 and HW3 follow the normal distribution with means equal to the original estimates of 3 hours and 2 hours, respectively, and each with a standard deviation of 0.25 hours. Use a chance constraint and Analytic Solver to find the solution that maximizes total profit while requiring at least a 95% chance of satisfying the original constraint for the third plant. d. Repeat part c, but now require at least a 99% chance of satisfying the original constraint at the third plant. AS 5.22.* Reconsider Problem 5.12, but now assume that the amount of resource 1 available and the amount of resource 2 available are both uncertain. In particular, each follows the normal distribution, with means equal to the original estimates of 8 and 4, respectively, but each with a standard deviation of 0.5. Use a chance constraint and Analytic Solver to find the maximum profit while requiring for each original constraint that there be at least a 95% chance of satisfying that constraint. AS 5.23. Reconsider Problem 5.13, but now assume that the available amount of resource 1 and available amount of resource 2 are both uncertain. a. If each unit of activity 1 is equally likely to use anywhere between 1.5 and 2.5 units of resource 3, while each unit of activity 2 is equally likely to use anywhere between 0.7 and 1.3 units of resource 3, then use a chance constraint and Analytic Solver to find the maximum profit while requiring at least a 95% chance of satisfying the original third constraint. b. If the amount of resource 3 used per unit of each activity follows the normal distribution, with means equal to the original estimates, but each with a standard deviation of 0.3, then use a chance constraint and Analytic Solver to find the maximum profit while requiring at least a 95% chance of satisfying the original third constraint. Case 5-1 Selling Soap 197 Case 5-1 Selling Soap Reconsider the Profit & Gambit Co. advertising-mix problem presented in Section 2.7. Recall that a major advertising campaign is being planned that will focus on three key products: a stain remover, a liquid detergent, and a powder detergent. Management has made the following policy decisions about what needs to be achieved by this campaign. • Sales of the stain remover should increase by at least 3 percent. • Sales of the liquid detergent should increase by at least 18 percent. • Sales of the powder detergent should increase by at least 4 percent. The spreadsheet in Figure 2.21 shows the linear programming model that was formulated for this problem. The minimum required increases in the sales of the three products are given in the data cells Minimum Increase (G8:G10). The changing cells Advertising Units (C14:D14) indicate that an optimal solution for the model is to undertake four units of advertising on television and three units of advertising in the print media. The objective cell TotalCost (G14) shows that the total cost for this advertising campaign would be $10 million. After receiving this information, Profit & Gambit management now wants to analyze the trade-off between the total advertising cost and the resulting benefits achieved by increasing the sales of the three products. Therefore, a management science team (you) has been given the assignment of developing the information that management will need to analyze this trade-off and decide whether it should change any of its policy decisions regarding the required minimum increases in the sales of the three products. In particular, management needs some detailed information about how the total advertising cost would change if it were to change any or all of these policy decisions. a. For each of the three products in turn, use graphical analysis to determine how much the total advertising cost would change if the required minimum increase in the sales of that product were to be increased by 1 percent (without changing the required minimum increases for the other two products). b. Use the spreadsheet shown in Figure 2.21 (available in www.mhhe.com/hillier6e) to obtain the information requested in part a. c. For each of the three products in turn, use a parameter analysis report to determine how the optimal solution for the model and the resulting total advertising cost would change if the required minimum increase in the sales of that product were to be systematically varied over a range of values (without changing the required minimum increases for the other two products). In each case, start the range of values at 0 percent and increase by 1 percent increments up to double the original minimum required increase. d. Use Solver to generate the sensitivity report and indicate how the report is able to provide the information requested in part a. Also use the report to obtain the allowable range for the required minimum increase in the sales of each product. Interpret how each of these allowable ranges relates to the results obtained in part c. e. Suppose that all the original numbers in Minimum Increase (G8:G10) were to be increased simultaneously by the same amount. How large can this amount be before the shadow prices provided by the sensitivity report may no longer be valid? f. Below is the beginning of a memorandum from the management science team to Profit & Gambit management that is intended to provide management with the information it needs to perform its trade-off analysis. Write the rest of this memorandum based on a summary of the results obtained in the preceding parts. Present your information in clear, simple terms that use the language of management. Avoid technical terms such as shadow prices, allowable ranges, and so forth. MEMORANDUM To: Profit & Gambit management From: The Management Science Team Subject: The trade-off between advertising expenditures and increased sales As instructed, we have been continuing our analysis of the plans for the major new advertising campaign that will focus on our spray prewash stain remover, our liquid formulation laundry detergent, and our powder laundry detergent. Our recent report presented our preliminary conclusions on how much advertising to do in the different media to meet the sales goals at a minimum total cost: Allocate $4 million to advertising on television. Allocate $6 million to advertising in the print media. Total advertising cost: $10 million. (Continued) 198 Chapter Five What-If Analysis for Linear Programming We estimate that the resulting increases in sales will be Stain remover: 3 percent increase in sales Liquid detergent: 18 percent increase in sales Powder detergent: 8 percent increase in sales You had specified that these increases should be at least 3 percent, 18 percent, and 4 percent, respectively, so we have met the minimum levels for the first two products and substantially exceeded it for the third. However, you also indicated that your decisions on these minimum required increases in sales (3 percent, 18 percent, and 4 percent) had been tentative ones. Now that we have more specific information on what the advertising costs and the resulting increases in sales will be, you plan to reevaluate these decisions to see if small changes might improve the trade-off between advertising cost and increased sales. To assist you in reevaluating your decisions, we now have analyzed this trade-off for each of the three products. Our best estimates are the following. Case 5-2 Controlling Air Pollution The Nori & Leets Co. is one of the major producers of steel in its part of the world. It is located in the city of Steeltown and is the only large employer there. Steeltown has grown and prospered along with the company, which now employs nearly 50,000 residents. Therefore, the attitude of the townspeople always has been, “What’s good for Nori & Leets is good for the town.” However, this attitude is now changing; uncontrolled air pollution from the company’s furnaces is ruining the appearance of the city and endangering the health of its residents. A recent stockholders’ revolt resulted in the election of a new enlightened board of directors for the company. These directors are determined to follow socially responsible policies, and they have been discussing with Steeltown city officials and citizens’ groups what to do about the air pollution problem. Together they have worked out stringent air quality standards for the Steeltown airshed. The three main types of pollutants in this airshed are particulate matter, sulfur oxides, and hydrocarbons. The new standards require that the company reduce its annual emission of these pollutants by the amounts shown in the following table. The board of directors has instructed management to have the engineering staff determine how to achieve these reductions in the most economical way. The steelworks have two primary sources of pollution, namely, the blast furnaces for making pig iron and the openhearth furnaces for changing iron into steel. In both cases, the engineers have decided that the most effective abatement methods are (1) increasing the height of the smokestacks,1 (2) using filter devices (including gas traps) in the smokestacks, and (3) including cleaner, high-grade materials among the fuels for the furnaces. Each of these methods has a technological limit on how heavily it can be used (e.g., a maximum feasible increase in the height of the smokestacks), but there also is considerable flexibility for using the method at a fraction of its technological limit. The table below shows how much emissions (in millions of pounds per year) can be eliminated from each type of furnace by fully using any abatement method to its technological limit. 1 Pollutant Particulates Sulfur oxides Hydrocarbons Required Reduction in Annual Emission Rate (million pounds) 60 150 125 Subsequent to this study, this particular abatement method has become a controversial one. Because its effect is to reduce ground-level pollution by spreading emissions over a greater distance, environmental groups contend that this creates more acid rain by keeping sulfur oxides in the air longer. Consequently, the U.S. Environmental Protection Agency adopted new rules to remove incentives for using tall smokestacks. Case 5-2 Controlling Air Pollution 199 Reduction in Emission Rate from the Maximum Feasible Use of an Abatement Method Taller Smokestacks Pollutant Particulates Sulfur oxides Hydrocarbons Filters Blast Furnaces Open-Hearth Furnaces Blast Furnaces Open-Hearth Furnaces Blast Furnaces Open-Hearth Furnaces 12 35 37 9 42 53 25 18 28 20 31 24 17 56 29 13 49 20 For purposes of analysis, it is assumed that each method also can be less fully used to achieve any fraction of the abatement capacities shown in this table. Furthermore, the fractions can be different for blast furnaces and open-hearth furnaces. For either type of furnace, the emission reduction achieved by each method is not substantially affected by whether or not the other methods also are used. After these data were developed, it became clear that no single method by itself could achieve all the required reductions. On the other hand, combining all three methods at full capacity on both types of furnaces (which would be prohibitively expensive if the company’s products are to remain competitively priced) is much more than adequate. Therefore, the engineers concluded that they would have to use some combination of the methods, perhaps with fractional capacities, based on their relative costs. Furthermore, because of the differences between the blast furnaces and the open-hearth furnaces, the best combinations may well be different for the two types of furnaces. An analysis was conducted to estimate the total annual cost that would be incurred by each abatement method. A method’s annual cost includes increased operating and maintenance expenses, as well as reduced revenue due to any loss in the efficiency of the production process caused by using the method. The other major cost is the start-up cost (the initial capital outlay) required to install the method. To make this one-time cost commensurable with the ongoing annual costs, the time value of money was used to calculate the annual expenditure that would be equivalent in value to this start-up cost. This analysis led to the total annual cost estimates given in the following table for using the methods at their full abatement capacities. Total Annual Cost from the Maximum Feasible Use of an Abatement Method Abatement Method Taller smokestacks Filters Better fuels Better Fuels Blast Furnaces Open-Hearth Furnaces $8 million 7 million 11 million $10 million 6 million 9 million It also was determined that the cost of a method being used at a lower level is roughly proportional to the fraction of the abatement capacity (given in the preceding table) that is achieved. Thus, for any given fraction achieved, the total annual cost would be roughly that fraction of the corresponding quantity in the cost table. The stage now is set to develop the general framework of the company’s plan for pollution abatement. This plan needs to specify which types of abatement methods will be used and at what fractions of their abatement capacities for (1) the blast furnaces and (2) the open-hearth furnaces. You have been asked to head a management science team to analyze this problem. Management wants you to begin by determining which plan would minimize the total annual cost of achieving the required reductions in annual emission rates for the three pollutants. a. Identify verbally the components of a linear programming model for this problem. b. Display the model on a spreadsheet. c. Obtain an optimal solution and generate the sensitivity report. Management now wants to conduct some what-if analysis with your help. Since the company does not have much prior experience with the pollution abatement methods under consideration, the cost estimates given in the third table are fairly rough, and each one could easily be off by as much as 10 percent in either direction. There also is some uncertainty about the values given in the second table, but less so than for the third table. By contrast, the values in the first table are policy standards and so are prescribed constants. However, there still is considerable debate about where to set these policy standards on the required reductions in the emission rates of the various pollutants. The numbers in the first table actually are preliminary values tentatively agreed upon before learning what the total cost would be to meet these standards. Both the city and company officials agree that the final decision on these policy standards should be based on the trade-off between costs and benefits. With this in mind, the city has concluded that each 10 percent increase in the policy standards over the current values (all the numbers in the first table) would be worth $3.5 million to the city. Therefore, the city has agreed to reduce the company’s tax payments to the city by $3.5 million for each 10 percent increase in the policy standards (up to 50 percent) that is accepted by the company. Finally, there has been some debate about the relative values of the policy standards for the three pollutants. As indicated in the first table, the required reduction for particulates now is less than half of that for either sulfur oxides or hydrocarbons. Some have argued for decreasing this disparity. Others contend that 200 Chapter Five What-If Analysis for Linear Programming an even greater disparity is justified because sulfur oxides and hydrocarbons cause considerably more damage than particulates. Agreement has been reached that this issue will be reexamined after information is obtained about which trade-offs in policy standards (increasing one while decreasing another) are available without increasing the total cost. d. Identify the parameters of the linear programming model that should be classified as sensitive parameters. Make a resulting recommendation about which parameters should be estimated more closely, if possible. e. Analyze the effect of an inaccuracy in estimating each cost parameter given in the third table. If the true value were 10 percent less than the estimated value, would this change the optimal solution? Would it change if the true value were 10 percent more than the estimated value? Make a resulting recommendation about where to focus further work in estimating the cost parameters more closely. f. For each pollutant, specify the rate at which the total cost of an optimal solution would change with any small change in the required reduction in the annual emission rate of the pollutant. Also specify how much this required reduction can be changed (up or down) without affecting the rate of change in the total cost. g. For each unit change in the policy standard for particulates given in the first table, determine the change in the opposite direction for sulfur oxides that would keep the total cost of an optimal solution unchanged. Repeat this for hydrocarbons instead of sulfur oxides. Then do it for a simultaneous and equal change for both sulfur oxides and hydrocarbons in the opposite direction from particulates. h. Letting θ denote the percentage increase in all the policy standards given in the first table, use a parameter analysis report to systematically find an optimal solution and the total cost for the revised linear programming problem for each θ = 10, 20, 30, 40, 50. Considering the tax incentive offered by the city, use these results to determine which value of θ (including the option of θ = 0) should be chosen by the company to minimize its total cost of both pollution abatement and taxes. i. For the value of θ chosen in part h, generate the sensitivity report and repeat parts f and g so that the decision makers can make a final decision on the relative values of the policy standards for the three pollutants. Case 5-3 Farm Management The Ploughman family owns and operates a 640-acre farm that has been in the family for several generations. The Ploughmans always have had to work hard to make a decent living from the farm and have had to endure some occasional difficult years. Stories about earlier generations overcoming hardships due to droughts, floods, and so forth, are an important part of the family history. However, the Ploughmans enjoy their self-reliant lifestyle and gain considerable satisfaction from continuing the family tradition of successfully living off the land during an era when many family farms are being abandoned or taken over by large agricultural corporations. John Ploughman is the current manager of the farm, while his wife Eunice runs the house and manages the farm’s finances. John’s father, Grandpa Ploughman, lives with them and still puts in many hours working on the farm. John and Eunice’s children, Frank, Phyllis, and Carl, also are given heavy chores before and after school. The entire family can produce a total of 4,000 person-hours’ worth of labor during the winter and spring months and 4,500 person-hours during the summer and fall. If any of these personhours are not needed, Frank, Phyllis, and Carl will use them to work on a neighboring farm for $7/hour during the winter and spring months and $7.70/hour during the summer and fall. The farm supports two types of livestock, dairy cows and laying hens, as well as three crops: soybeans, corn, and wheat. (All three are cash crops, but the corn also is a feed crop for the cows and the wheat also is used for chicken feed.) The crops are harvested during the late summer and fall. During the winter months, John, Eunice, and Grandpa make a decision about the mix of livestock and crops for the coming year. Currently, the family has just completed a particularly successful harvest that has provided an investment fund of $28,000 that can be used to purchase more livestock. (Other money is available for ongoing expenses, including the next planting of crops.) The family currently has 30 cows valued at $49,000 and 2,000 hens valued at $7,000. They wish to keep all this livestock and perhaps purchase more. Each new cow would cost $2,100, and each new hen would cost $4.20. Over a year’s time, the value of a herd of cows will decrease by about 10 percent and the value of a flock of hens will decrease by about 25 percent due to aging. Each cow will require two acres of land for grazing and 10 person-hours of work per month, while producing a net annual cash income of $1,190 for the family. The corresponding figures for each hen are no significant acreage, 0.05 person-hours per month, and an annual net cash income of $5.95. The chicken house can accommodate a maximum of 5,000 hens, and the size of the barn limits the herd to a maximum of 42 cows. For each acre planted in each of the three crops, the first table on the next page gives the number of person-hours of work that will be required during the first and second halves of the year, as well as a rough estimate of the crop’s net value (in either income or savings in purchasing feed for the livestock). To provide much of the feed for the livestock, John wants to plant at least one acre of corn for each cow in the coming year’s Case 5-3 Farm Management 201 herd and at least 0.05 acre of wheat for each hen in the coming year’s flock. John, Eunice, and Grandpa now are discussing how much acreage should be planted in each of the crops and how many cows and hens to have for the coming year. Their objective is to maximize the family’s monetary worth at the end of the coming year (the sum of the net income from the livestock for the coming year plus the net value of the crops for the coming year plus what remains from the investment fund plus the value of the livestock at the end of the coming year plus income from working on a neighboring farm minus living expenses of $56,000 for the year). Soybeans Corn Wheat Winter and spring, person-hours 1.0 0.9 0.6 Summer and fall, person-hours 1.4 1.2 0.7 $98 $84 $56 Net value a. Identify verbally the components of a linear programming model for this problem. b. Display the model on a spreadsheet. c. Obtain an optimal solution and generate the sensitivity report. What does the model predict regarding the family’s monetary worth at the end of the coming year? d. Find the allowable range for the net value per acre planted for each of the three crops. The above estimates of the net value per acre planted in each of the three crops assumes good weather conditions. Adverse weather conditions would harm the crops and greatly reduce the resulting value. The scenarios particularly feared by the family are a drought, a flood, an early frost, both a drought and an early frost, and both a flood and an early frost. The estimated net values for the year under these scenarios are shown next. Net Value per Acre Planted Scenario Drought Flood Early frost Drought and early frost Flood and early frost Soybeans Corn Wheat −$14 −$21 0 21 28 $14 70 56 42 −21 −28 −14 14 14 7 e. Find an optimal solution under each scenario after making the necessary adjustments to the linear programming model formulated in part b. In each case, what is the prediction regarding the family’s monetary worth at the end of the year? f. For the optimal solution obtained under each of the six scenarios (including the good weather scenario considered in parts a-d), calculate what the family’s monetary worth would be at the end of the year if each of the other five scenarios occurs instead. In your judgment, which solution provides the best balance between yielding a large monetary worth under good weather conditions and avoiding an overly small monetary worth under adverse weather conditions? Grandpa has researched what the weather conditions were in past years as far back as weather records have been kept and obtained the data shown in the table below. With these data, the family has decided to use the following approach to making its planting and livestock decisions. Rather than the optimistic approach of assuming that good weather conditions will prevail (as done in parts a-d), the average net value under all weather conditions will be used for each crop (weighting the net values under the various scenarios by the frequencies in the following table). Scenario Frequency Good weather 40% Drought 20 Flood 10 Early frost 15 Drought and early frost 10 Flood and early frost 5 g. Modify the linear programming model formulated in part b to fit this new approach. h. Repeat part c for this modified model. i. Use a shadow price obtained in part h to analyze whether it would be worthwhile for the family to obtain a bank loan with a 10 percent interest rate to purchase more livestock now beyond what can be obtained with the $28,000 from the investment fund. j. For each of the three crops, use the sensitivity report obtained in part h to identify how much latitude for error is available in estimating the net value per acre planted for that crop without changing the optimal solution. Which two net values need to be estimated most carefully? If both estimates are incorrect simultaneously, how close do the estimates need to be to guarantee that the optimal solution will not change? Use a two-way parameter analysis report to systematically generate the optimal monetary worth as these two net values are varied simultaneously over ranges that go up to twice as far from the estimates as needed to guarantee that the optimal solution will not change. This problem illustrates a kind of situation that is frequently faced by various kinds of organizations. To describe the situation in general terms, an organization faces an uncertain future where any one of a number of scenarios may unfold. Which one will occur depends on conditions that are outside the control of the organization. The organization needs to choose the levels of various activities, but the unit contribution of each activity to the overall measure of performance is greatly affected by which scenario unfolds. Under these circumstances, what is the best mix of activities? k. Think about specific situations outside of farm management that fit this description. Describe one. 202 Chapter Five What-If Analysis for Linear Programming Case 5-4 Assigning Students to Schools (Revisited) Reconsider Case 3-5. The Springfield School Board still has the policy of providing busing for all middle school students who must travel more than approximately a mile. Another current policy is to allow splitting residential areas among multiple schools if this will reduce the total busing cost. (This latter policy will be reversed in Case 7-3.) However, before adopting a busing plan based on part a of Case 3-5, the school board now wants to conduct some what-if analysis. a. If you have not already done so for part a of Case 3-5, formulate and solve a linear programming model for this problem on a spreadsheet. b. Use Solver to generate the sensitivity report. One concern of the school board is the ongoing road construction in area 6. These construction projects have been delaying traffic considerably and are likely to affect the cost of busing students from area 6, perhaps increasing costs as much as 10 percent. c. Use the sensitivity report to check how much the busing cost from area 6 to school 1 can increase (assuming no change in the costs for the other schools) before the current optimal solution would no longer be optimal. If the allowable increase is less than 10 percent, use Solver to find the new optimal solution with a 10 percent increase. d. Repeat part c for school 2 (assuming no change in the costs for the other schools). e. Now assume that the busing cost from area 6 would increase by the same percentage for all the schools. Use the sensitivity report to determine how large this percentage can be before the current optimal solution might no longer be optimal. If the allowable increase is less than 10 percent, use Solver to find the new optimal solution with a 10 percent increase. The school board has the option of adding portable classrooms to increase the capacity of one or more of the middle schools for a few years. However, this is a costly move that the board would only consider if it would significantly decrease busing costs. Each portable classroom holds 20 students and has a leasing cost of $2,500 per year. To analyze this option, the school board decides to assume that the road construction in area 6 will wind down without significantly increasing the busing costs from that area. f. For each school, use the corresponding shadow price from the sensitivity report to determine whether it would be worthwhile to add any portable classrooms. g. For each school where it is worthwhile to add any portable classrooms, use the sensitivity report to determine how many could be added before the shadow price would no longer be valid (assuming this is the only school receiving portable classrooms). h. If it would be worthwhile to add portable classrooms to more than one school, use the sensitivity report to determine the combinations of the number to add for which the shadow prices definitely would still be valid. Then use the shadow prices to determine which of these combinations is best in terms of minimizing the total cost of busing students and leasing portable classrooms. Use Solver for finding the corresponding optimal solution for assigning students to schools. i. If part h was applicable, modify the best combination of portable classrooms found there by adding one more to the school with the most favorable shadow price. Use Solver to find the corresponding optimal solution for assigning students to schools and to generate the corresponding sensitivity report. Use this information to assess whether the plan developed in part h is the best one available for minimizing the total cost of busing students and leasing portables. If not, find the best plan. Additional Case Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases.ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. Chapter SIX Network Optimization Problems Learning Objectives After completing this chapter, you should be able to 1. Formulate network models for various types of network optimization problems. 2. Describe the characteristics of minimum-cost flow problems, maximum flow problems, and shortest path problems. 3. Identify some areas of application for these types of problems. 4. Identify several categories of network optimization problems that are special types of minimum-cost flow problems. 5. Formulate and solve a spreadsheet model for a minimum-cost flow problem, a maximum flow problem, or a shortest path problem from a description of the problem. This chapter focuses on how to analyze networks, so let's begin with a description of what they are. A network can be defined as a system with a number of components where there are direct connections between various pairs of components. These connections might be actual physical connections, such as the wires in an electrical network. In other cases, the network might be a diagram that depicts the locations of the components of the system and then uses lines between certain pairs of components to identify direct relationships between these components. For example, if you refer back to Section 3.5, Figure 3.9 shows a network that depicts a distribution system with two types of components (factories and customers) where the lines from factories to customers identify the feasible shipping routes. Networks arise in numerous settings and in a variety of guises. Transportation, electrical, and communication networks pervade our daily lives. Network representations also are widely used for problems in such diverse areas as production, distribution, project planning, facilities location, resource management, and financial planning—to name just a few examples. In fact, a network representation provides such a powerful visual and conceptual aid for portraying the relationships between the components of systems that it is used in virtually every field of scientific, social, and economic endeavor. This chapter focuses on how to use spreadsheet models to optimize the operation of certain kinds of networks. All of these networks are designed to provide flow of some type from certain points of origin along various permissible routes to certain destinations. For example, this flow might involve the shipment of goods or resources through the network. Or perhaps the objective is to identify the optimal route through the network, in which case the flow simply involves travel along that route. For each type of network, we will show how a spreadsheet model can formulate and solve the problem of how to operate the network in an optimal manner. One of the most exciting developments in management science in recent decades has been the unusually rapid advance in both the methodology and application of network optimization problems. A number of algorithmic breakthroughs have had a major impact, as have ideas from computer science concerning data structures and efficient data manipulation. Consequently, algorithms and software now are available and are being used to solve huge problems on a routine basis that would have been completely intractable a few decades ago. However, we will only consider small examples in this chapter. 203 204 Chapter Six Network Optimization Problems This chapter presents the network optimization problems that have been particularly helpful in dealing with managerial issues. We focus on the nature of these problems and their applications rather than on the technical details and the algorithms used to solve the problems. Using spreadsheet models and Solver will enable us to maintain this focus. You already have seen some examples of network optimization problems in Chapter 3. In particular, transportation problems (described in Section 3.5) have a network representation that is shown in Figure 3.9 already cited earlier. A typical transportation problem involves shipping some product from certain factories to certain customers, where the objective is to identify the shipping plan that minimizes the total cost of these shipments. Assignment problems (Section 3.6) have a similar network representation (as described further in Chapter 15 at www.mhhe.com/Hillier6e). Therefore, both transportation problems and assignment problems are simple types of network optimization problems. Like transportation problems and assignment problems, many other network optimization problems (including all the types considered in this chapter) also are special types of linear programming problems. Consequently, after formulating a spreadsheet model for these problems, they can be readily solved by Solver. Section 6.1 discusses an especially important type of network optimization problem called a minimum-cost flow problem. A typical application involves minimizing the cost of shipping goods through a distribution network. Thus, this problem is similar to a transportation problem except now there are some intermediate points (e.g., warehouses) in the distribution network. Section 6.3 presents maximum flow problems, which are concerned with such issues as how to maximize the flow of goods through a distribution network. Section 6.2 lays the groundwork by introducing a case study of a maximum flow problem. Section 6.4 considers shortest path problems. In their simplest form, the objective is to find the shortest route through a network between two locations. A supplement to this chapter at www.mhhe.com/Hillier6e discusses minimum spanningtree problems, which are concerned with minimizing the cost of providing either direct or indirect connections between all users of a system. This is the only network optimization problem considered in this book that is not, in fact, a special type of linear programming problem. 6.1 MINIMUM-COST FLOW PROBLEMS Before describing the general characteristics of minimum-cost flow problems, let us first look at an example. For clarity, this first example is a tiny one, but otherwise it is typical of the sometimes huge minimum-cost flow problems that arise in practice. An Example: The Distribution Unlimited Co. Problem The Distribution Unlimited Co. has two factories producing a product that needs to be shipped to two warehouses. Here are some details. Factory 1 is producing 80 units. Factory 2 is producing 70 units. Warehouse 1 needs 60 units. Warehouse 2 needs 90 units. The objective is to minimize the total shipping cost through the distribution network. (Each unit corresponds to a full truckload of the product.) Figure 6.1 shows the distribution network available for shipping this product, where F1 and F2 are the two factories, W1 and W2 are the two warehouses, and DC is a distribution center. The arrows show feasible shipping lanes. In particular, there is a rail link from Factory 1 to Warehouse 1 and another from Factory 2 to Warehouse 2. (Any amounts can be shipped along these rail links.) In addition, independent truckers are available to ship up to 50 units from each factory to the distribution center, and then to ship up to 50 units from the distribution center to each warehouse. (Whatever is shipped to the distribution center must subsequently be shipped on to the warehouses.) Management’s objective is to determine the shipping plan (how many units to ship along each shipping lane) that will minimize the total shipping cost. 6.1 Minimum-Cost Flow Problems 205 FIGURE 6.1 The distribution network for the Distribution Unlimited Co. problem, where each feasible shipping lane is represented by an arrow. 80 units produced F1 W1 60 units needed W2 90 units needed DC 70 units produced F2 If you have read the description of transportation problems in Section 3.5 (as illustrated in Figure 3.9), you will notice that this minimum-cost flow problem resembles a transportation problem. However, the key difference is that this example includes an intermediate destination (the distribution center), which does not fit the definition of a transportation problem. Many applications include such intermediate destinations, which makes minimum-cost flow problems a valuable generalization of transportation problems. The shipping costs differ considerably among these shipping lanes. The cost per unit shipped through each lane is shown above the corresponding arrow in the network in Figure 6.2. To make the network less crowded, the problem usually is presented even more compactly, as shown in Figure 6.3. The number in square brackets next to the location of each facility indicates the net number of units (outflow minus inflow) generated there. Thus, the number of units terminating at each warehouse is shown as a negative number. The number at the distribution center is 0 since the number of units leaving minus the number of units arriving must equal 0. The number on top of each arrow shows the unit shipping cost along that shipping lane. Any number in square brackets underneath an arrow gives [80] $700/unit 60 units needed $3 DC [5 0 0 t ni x.] a /u 00 s m it un [5 [0] $2 0 un 0/u its nit m ax .] W1 00 $3 0] [5 F1 0 [5 F2 FIGURE 6.2 [5 t ni x.] a /u m 00 $4 nits u 0 [5 $5 0 un 0/u its nit m ax .] 0] $5 00 00 $4 0] [5 DC 70 units produced W1 $2 [5 00 0] F1 80 units produced [–60] $700 $1,000/unit F2 W2 90 units needed The data for the distribution network for the ­Distribution Unlimited Co. problem. [70] FIGURE 6.3 $1,000 W2 [–90] A network model for the Distribution Unlimited Co. problem as a minimum-cost flow problem. An Application Vignette Hewlett-Packard (HP) offers many innovative products to meet the diverse needs of more than 1 billion customers. The breadth of its product offering has helped the company achieve unparalleled market reach. However, offering multiple similar products also can cause serious problems—including confusing sales representatives and customers—that can adversely affect the revenue and costs for any particular product. Therefore, it is important to find the right balance between too much product variety and too little. With this in mind, HP top management made managing product variety a strategic business priority. HP has been a leader in applying management science to its important business problems for decades, so it was only natural that many of the company’s top management scientists were called on to address this problem as well. The heart of the methodology that was developed to address this problem involved formulating and applying a network optimization model. After excluding proposed products that do not have a sufficiently high return on investment, the remaining Figure 6.3 illustrates how a minimum-cost flow problem can be completely depicted by a network. proposed products can be envisioned as flows through a network that can help fill some of the projected orders on the righthand side of the network. The resulting model is a special type of minimum cost flow problem (related to the special type discussed in the next two sections). Following its implementation by the beginning of 2005, this application of a minimum cost flow problem had a dramatic impact in enabling HP businesses to increase operational focus on their most critical products. This yielded companywide profit improvements of over $500 million between 2005 and 2008, and then about $180 million annually thereafter. It also yielded a variety of important qualitative benefits for HP. These dramatic results led to HP winning the prestigious first prize in the 2009 Franz Edelman Award for Achievement in Operations Research and the Management Sciences. Source: J. Ward and 20 co-authors, “HP Transforms Product Portfolio Management with Operations Research,” Interfaces 40, no. 1 (January–February 2010), pp. 17–32. (A link to this article is provided at www.mhhe.com/Hillier6e.) the maximum number of units that can be shipped along that shipping lane. (The absence of a number in square brackets underneath an arrow implies that there is no limit on the shipping amount there.) This network provides a complete representation of the problem, including all the necessary data, so it constitutes a network model for this minimum-cost flow problem. Since this is such a tiny problem, you probably can see what the optimal solution must be. (Try it.) This solution is shown in Figure 6.4, where the shipping amount along each shipping lane is given in parentheses. (To avoid confusion, we delete the unit shipping costs and shipping capacities in this figure.) Combining these shipping amounts with the unit shipping costs given in Figures 6.2 and 6.3, the total shipping cost for this solution (when starting by listing the costs from F1, then from F2, and then from DC) is Total shipping cost = 30($700) + 50($300) + 30($500) + 40($1,000) + 30($200) + 50($400) = $117,000 [80] [–60] (30) F1 W1 0) 0) (5 The optimal solution for the Distribution Unlimited Co. problem, where the shipping amounts are shown in parentheses over the arrows. (3 FIGURE 6.4 [0] DC (3 0) 0) (5 F2 [70] 206 (40) W2 [–90] 6.1 Minimum-Cost Flow Problems 207 General Characteristics This example possesses all the general characteristics of any minimum-cost flow problem. Before summarizing these characteristics, here is the terminology you will need. Terminology A supply node has net flow going out whereas a demand node has net flow coming in. 1. The model for any minimum-cost flow problem is represented by a network with flow passing through it. 2. The circles in the network are called nodes. 3. Each node where the net amount of flow generated (outflow minus inflow) is a fixed ­positive number is a supply node. (Thus, F1 and F2 are the supply nodes in Figure 6.3.) 4. Each node where the net amount of flow generated is a fixed negative number is a demand node. (Consequently, W1 and W2 are the demand nodes in the example.) 5. Any node where the net amount of flow generated is fixed at zero is a transshipment node. (Thus, DC is the transshipment node in the example.) Having the amount of flow out of the node equal the amount of flow into the node is referred to as conservation of flow. 6. The arrows in the network are called arcs. 7. The maximum amount of flow allowed through an arc is referred to as the capacity of that arc. Using this terminology, the general characteristics of minimum-cost flow problems (the model for this type of problem) can be described in terms of the following assumptions. Assumptions of a Minimum-Cost Flow Problem Since the arrowhead on an arc indicates the direction in which flow is allowed, a pair of arcs pointing in opposite directions is used if flow can occur in both directions. The objective is to minimize the total cost of supplying the demand nodes. 1. At least one of the nodes is a supply node, so each one specifies its own fixed positive ­number for the net amount of flow generated there. 2. At least one of the other nodes is a demand node, so each one specifies its own fixed ­negative number for the net amount of flow generated there. 3. All the remaining nodes are transshipment nodes, so each one specifies a fixed value of zero for the net amount of flow generated there. 4. Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. (If flow can occur in both directions, this would be represented by a pair of arcs pointing in opposite directions.) 5. The network has enough arcs with sufficient capacity to enable all the flow generated at the supply nodes to reach all the demand nodes. 6. The cost of the flow through each arc is proportional to the amount of that flow, where the cost per unit flow is known. 7. The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. (An alternative objective is to maximize the total profit from doing this.) A solution for this kind of problem needs to specify how much flow is going through each arc. To be a feasible solution, the amount of flow through each arc cannot exceed the capacity of that arc and the net amount of flow generated at each node must equal the specified amount for that node. The following property indicates when the problem will have feasible solutions. Feasible Solutions Property: Under the above assumptions, a minimum-cost flow problem will have feasible solutions if and only if the sum of the supplies from its supply nodes equals the sum of the demands at its demand nodes. Note that this property holds for the Distribution Unlimited Co. problem, because the sum of its supplies is 80 + 70 = 150 and the sum of its demands is 60 + 90 = 150. Applications occasionally do arise where the sum of the supplies from the supply nodes does not equal the sum of the demands at its demand nodes (an “unbalanced problem”). For 208 Chapter Six Network Optimization Problems example, suppose that the sum of the supplies exceeds the sum of the demands by some amount (the “extra supply”). In this case, the “supplies” actually represent the maximum amounts that could be supplied rather than fixed amounts, so part of the problem is to determine exactly how much should be produced at each supply node to exactly meet the demands at the demand nodes. Such a problem no longer exactly fits the definition of a minimumcost flow problem because it violates assumption 1. Nevertheless, Solver can still solve the problem by using the corresponding spreadsheet model that includes ≤ constraints for the amounts produced at each supply node. Alternatively, the problem can be converted back into a minimum-cost flow problem by adding a “dummy demand node” to the network where the demand at this node equals the “extra supply” at the supply nodes. By adding an arc with a zero cost per unit flow (and a large arc capacity) from each supply node to the dummy demand node, the network now becomes a minimum-cost flow problem because it now satisfies the feasible solutions property. For many applications of minimum-cost flow problems, management desires a solution with integer values for all the flow quantities (e.g., integer numbers of full truckloads along each shipping lane). The model does not include any constraints that require this for feasible solutions. Fortunately, such constraints are not needed because of the following property. Integer Solutions Property: As long as all its supplies, demands, and arc capacities have integer values, any minimum-cost flow problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its flow quantities. See in Figure 6.3 that all the assumptions needed for this property to hold are satisfied for the Distribution Unlimited Co. problem. In particular, all the supplies (80 and 70), demands (60 and 90), and arc capacities (50) have integer values. Therefore, all the flow quantities in the optimal solution given in Figure 6.4 (30 three times, 50 two times, and 40) have integer values. This ensures that only full truckloads will be shipped into and out of the distribution center. (Remember that each unit corresponds to a full truckload of the product.) Now let us see how to obtain an optimal solution for the Distribution Unlimited Co. problem by formulating a spreadsheet model and then applying Solver. Using Excel to Formulate and Solve Minimum-Cost Flow Problems Figure 6.5 shows a spreadsheet model that is based directly on the network representation of the problem in Figure 6.3. The arcs are listed in columns B and C, along with their capacities (unless unlimited) in column F and their costs per unit flow in column G. The changing cells Ship (D4:D9) show the flow amounts through these arcs and the objective cell TotalCost (D11) provides the total cost of this flow by using the equation D11 = SUMPRODUCT(Ship, UnitCost) Capacity constraints like these are needed in any minimum-cost flow problem that has any arcs with limited capacity. Any minimum-cost flow problem needs net flow constraints like this for every node. Excel Tip: SUMIF(A, B, C) adds up each entry in the range C for which the corresponding entry in range A equals B. This function is especially useful in network problems for calculating the net flow generated at a node. The first set of constraints in the Solver Parameters box, D5:D8 ≤ Capacity (F5:F8), ensures that the arc capacities are not exceeded. Similarly, Column I lists the nodes, column J calculates the actual net flow generated at each node (given the flows in the changing cells), and column L specifies the net amount of flow that needs to be generated at each node. Thus, the second set of constraints in the Solver Parameters box is NetFlow (J4:J8) = SupplyDemand (L4:L8), requiring that the actual net amount of flow generated at each node must equal the specified amount. The equations entered into NetFlow (J4:J8) use the difference of two SUMIF functions to calculate the net flow (outflow minus inflow) generated at each node. In particular, the first SUMIF function calculates the flow leaving the node and the second one calculates the flow entering the node. (If you haven't used SUMIF functions previously, the nearby Excel tip defines what this kind of function does.) For example, consider the F1 node (I4) and the calculation in J4 of the net flow for this node. The equation for this net flow is J4 = SUMIF(From, I4, Ship) − SUMIF(To, I4, Ship) 6.1 Minimum-Cost Flow Problems 209 FIGURE 6.5 A spreadsheet model for the Distribution Unlimited Co. minimum-cost flow problem, including the objective cell TotalCost (D11) and the other output cells NetFlow (J4:J8), as well as the equations entered into these cells and the other specifications needed to set up the model. The changing cells Ship (D4:D9) show the optimal shipping quantities through the distribution network obtained by Solver. A 1 B C D E F G H K L I J Unit Cost Nodes Net Flow $700 F1 80 = 80 70 Distribution Unlimited Co. Minimum Cost Flow Problem 2 3 From To Ship 4 F1 W1 30 Capacity Supply/Demand 5 F1 DC 50 ≤ 50 $300 F2 70 = 6 DC W1 30 ≤ 50 $200 DC 0 = 0 7 DC W2 50 ≤ 50 $400 W1 -60 = -60 ≤ 50 $500 W2 -90 = -90 8 F2 DC 30 9 F2 W2 40 $1,000 10 11 Total Cost $117,000 Solver Parameters Set Objective Cell: TotalCost To: Min By Changing Variable Cells: Ship Subject to the Constraints: D5:D8 <= Capacity NetFlow = SupplyDemand Solver Options: Make Variables Nonnegative Solving Method: Simplex LP J Range Name Cells Capacity From NetFlow Nodes Ship SupplyDemand To TotalCost UnitCost F5:F8 B4:B9 J4:J8 I4:I8 D4:D9 L4:L8 C4:C9 D11 G4:G9 11 3 Net Flow 4 =SUMIF(From,I4,Ship)-SUMIF(To,I4,Ship) 5 =SUMIF(From,I5,Ship)-SUMIF(To,I5,Ship) 6 =SUMIF(From,I6,Ship)-SUMIF(To,I6,Ship) 7 =SUMIF(From,I7,Ship)-SUMIF(To,I7,Ship) 8 =SUMIF(From,I8,Ship)-SUMIF(To,I8,Ship) C D Total Cost =SUMPRODUCT(Ship,UnitCost) SUMIF(From,I4,Ship) sums each individual entry in Ship (D4:D9) if that entry is in a row where the entry in From (B4:B9) is the same as in I4. Since I4 = F1 and the only rows that have F1 in the From column are rows 4 and 5, the sum in the Ship column is only over these same rows, so this sum is D4 + D5. Similarly, SUMIF(To,I4,Ship) sums each individual entry in Ship (D4:D9) if that entry is in a row where the entry in To (C4:C9) is the same as in I4. However, F1 never appears in the To column, so this sum is 0. Therefore, the overall equation for J4 yields J4 = D4 + D5 − 0 = 30 + 50 − 0 = 80, which is the net flow generated at the F1 node. While it appears more complicated to use the SUMIF function rather than just entering J4 = D4 + D5, J5 = D8 + D9, J6 = D6 + D7 − D5 − D8, and so on, it is actually simpler. The SUMIF formula only needs to be entered once (in cell J4). It can then be copied down into the remaining cells in NetFlow (J5:J8). For a problem with many nodes, this is much quicker and (perhaps more significantly) less prone to error. In a large problem, it is all too easy to miss an arc when determining which cells in the Ship column to add and subtract to calculate the net flow for a given node. The first Solver option specifies that the flow amounts cannot be negative. The second acknowledges that this is still a linear programming problem. Running Solver gives the optimal solution shown in Ship (D4:D9) in Figure 6.5. This is the same solution as displayed in Figure 6.4. 210 Chapter Six Network Optimization Problems Solving Large Minimum-Cost Flow Problems More Efficiently network simplex method The network simplex method can solve much larger minimum-cost flow problems (sometimes with millions of nodes and arcs) than can the simplex method used by Solver. Because minimum-cost flow problems are a special type of linear programming problem, and the simplex method can solve any linear programming problem, it also can solve any minimumcost flow problem in the standard way. For example, Solver uses the simplex method to solve this type (or any other type) of linear programming problem. This works fine for small problems, like the Distribution Unlimited Co. problem, and for considerably larger ones as well. Therefore, the approach illustrated in Figure 6.5 will serve you well for any minimum-cost flow problem encountered in this book and for many that you will encounter subsequently. However, we should mention that a different approach is sometimes needed in practice to solve really big problems. Because of the special form of minimum-cost flow problems, it is possible to greatly streamline the simplex method to solve them far more quickly. In particular, rather than going through all the algebra of the simplex method, it is possible to execute the same steps far more quickly by working directly with the network for the problem. This streamlined version of the simplex method is called the network simplex method. The network simplex method can solve some huge problems that are much too large for the simplex method. Like the simplex method, the network simplex method not only finds an optimal solution but also can be a valuable aid to managers in conducting the kinds of what-if analyses described in Chapter 5. Many companies now use the network simplex method to solve their minimum-cost flow problems. Some of these problems are huge, with many tens of thousands of nodes and arcs. Occasionally, the number of arcs will even be far larger, perhaps into the millions. Although Solver does not, other commercial software packages for linear programming commonly include the network simplex method. Another important advance in fairly recent years has been the development of excellent graphical interfaces for modeling minimum-cost flow problems. These interfaces make the design of the model and the interpretation of the output of the network simplex method completely visual and intuitive with no mathematics involved. This is very helpful for managerial decision making. Some Applications Probably the most important kind of application of minimum-cost flow problems is to the operation of a distribution network, such as the one depicted in Figures 6.1–6.4 for the Distribution Unlimited Co. problem. As summarized in the first row of Table 6.1, this kind of application involves determining a plan for shipping goods from their sources (factories, etc.) to intermediate storage facilities (as needed) and then on to the customers. For some applications of minimum-cost flow problems, all the transshipment nodes are processing facilities rather than intermediate storage facilities. This is the case for solid waste management, as indicated in Table 6.1. Here, the flow of materials through the network begins at the sources of the solid waste, then goes to the facilities for processing these waste materials into a form suitable for landfill, and then sends them on to the various landfill locations. However, the objective still is to determine the flow plan that minimizes the total cost, where the cost now is for both shipping and processing. TABLE 6.1 Typical Kinds of Applications of Minimum-Cost Flow Problems Kind of Application Supply Nodes Transshipment Nodes Demand Nodes Operation of a distribution network Solid waste management Operation of a supply network Sources of goods Sources of solid waste Vendors Intermediate storage facilities Processing facilities Intermediate warehouses Coordinating product mixes at plants Plants Sources of cash at a specific time Production of a specific product Customers Landfill locations Processing facilities Market for a specific product Needs for cash at a specific time Cash flow management Short-term investment options 6.1 Minimum-Cost Flow Problems 211 In other applications, the demand nodes might be processing facilities. For example, in the third row of Table 6.1, the objective is to find the minimum-cost plan for obtaining supplies from various possible vendors, storing these goods in warehouses (as needed), and then shipping the supplies to the company’s processing facilities (factories, etc.). The next kind of application in Table 6.1 (coordinating product mixes at plants) illustrates that arcs can represent something other than a shipping lane for a physical flow of materials. This application involves a company with several plants (the supply nodes) that can produce the same products but at different costs. Each arc from a supply node represents the production of one of the possible products at that plant, where this arc leads to the transshipment node that corresponds to this product. Thus, this transshipment node has an arc coming in from each plant capable of producing this product, and then the arcs leading out of this node go to the respective customers (the demand nodes) for this product. The objective is to determine how to divide each plant’s production capacity among the products so as to minimize the total cost of meeting the demand for the various products. The last application in Table 6.1 (cash flow management) illustrates that different nodes can represent some event that occurs at different times. In this case, each supply node represents a specific time (or time period) when some cash will become available to the company (through maturing accounts, notes receivable, sales of securities, borrowing, etc.). The supply at each of these nodes is the amount of cash that will become available then. Similarly, each demand node represents a specific time (or time period) when the company will need to draw on its cash reserves. The demand at each such node is the amount of cash that will be needed then. The objective is to maximize the company’s income from investing the cash between each time it becomes available and when it will be used. Therefore, each transshipment node represents the choice of a specific short-term investment option (e.g., purchasing a certificate of deposit from a bank) over a specific time interval. The resulting network will have a succession of flows representing a schedule for cash becoming available, being invested, and then being used after the maturing of the investment. Special Types of Minimum-Cost Flow Problems transshipment problem A transshipment problem is just a minimum-cost flow problem that has unlimited capacities for all its arcs. There are five important categories of network problems that turn out to be special types of minimum-cost flow problems. One is the transportation problems discussed in Section 3.5.Recall that a typical kind of transportation problem involves minimizing the cost of shipping something (e.g., many units of a particular product) from a number of sources (e.g., plants producing this product) directly to a number of destinations (e.g., customers). Figure 3.9 shows the network representation of a typical transportation problem. In our current terminology, the sources and destinations of a transportation problem are the supply nodes and demand nodes, respectively. Thus, a transportation problem is just a minimum-cost flow problem without any transshipment nodes and without any capacity constraints on the arcs (all of which go directly from a supply node to a demand node). A second category is the assignment problems discussed in Section 3.6. Recall that this kind of problem involves assigning a group of people (or other operational units) to a group of tasks where each person is to perform a single task. An assignment problem can be viewed as a special type of transportation problem whose sources are the assignees and whose destinations are the tasks. This then makes the assignment problem also a special type of minimum-cost flow problem with the characteristics described in the preceding paragraph. In addition, each person is a supply node with a supply of 1 and each task is a demand node with a demand of 1. A third special type of minimum-cost flow problem is transshipment problems. This kind of problem is just like a transportation problem except for the additional feature that the shipments from the sources (supply nodes) to the destinations (demand nodes) might also pass through intermediate transfer points (transshipment nodes) such as distribution centers. Like a transportation problem, there are no capacity constraints on the arcs. Consequently, any minimum-cost flow problem where each arc can carry any desired amount of 212 Chapter Six Network Optimization Problems The network simplex method can be used to solve huge problems of any of these five special types. Review Questions 6.2 flow is a transshipment problem. For example, if the data in Figure 6.2 were altered so that any amounts (within the ranges of the supplies and demands) could be shipped into and out of the distribution center, the Distribution Unlimited Co. would become just a transshipment problem.1 Because of their close relationship to a general minimum-cost flow problem, we will not discuss transshipment problems further. The other two important special types of minimum-cost flow problems are maximum flow problems and shortest path problems, which will be described in Sections 6.3 and 6.4 after presenting a case study of a maximum flow problem in the next section. In case you are wondering why we are bothering to point out that these five kinds of problems are special types of minimum-cost flow problems, here is one very important reason. It means that the network simplex method can be used to solve huge problems of any of these types that might be difficult or impossible for the simplex method to solve. It is true that other efficient special-purpose algorithms also are available for each of these kinds of problems. However, recent implementations of the network simplex method have become so powerful that it now provides an excellent alternative to these other algorithms in most cases. This is especially valuable when the available software package includes the network simplex method but not another relevant special-purpose algorithm. Furthermore, even after finding an optimal solution, the network simplex method can continue to be helpful in aiding managerial what-if sessions along the lines discussed in Chapter 5. 1. 2. 3. 4. 5. 6. Name and describe the three kinds of nodes in a minimum-cost flow problem. What is meant by the capacity of an arc? What is the usual objective for a minimum-cost flow problem? What property is necessary for a minimum-cost flow problem to have feasible solutions? What is the integer solutions property for minimum-cost flow problems? What is the name of the streamlined version of the simplex method that is designed to solve minimum-cost flow problems very efficiently? 7. What are a few typical kinds of applications of minimum-cost flow problems? 8. Name five important categories of network optimization problems that turn out to be s­ pecial types of minimum-cost flow problems. A CASE STUDY: THE BMZ CO. MAXIMUM FLOW PROBLEM What a day! First being called into his boss’s office and then receiving an urgent telephone call from the company president himself. Fortunately, he was able to reassure them that he has the situation under control. Although his official title is Supply Chain Manager for the BMZ Company, Karl Schmidt often tells his friends that he really is the company’s crisis manager. One crisis after another. The supplies needed to keep the production lines going haven’t arrived yet. Or the supplies have arrived but are unusable because they are the wrong size. Or an urgent shipment to a key customer has been delayed. This current crisis is typical. One of the company’s most important distribution centers—the one in Los Angeles—urgently needs an increased flow of shipments from the company. Karl was chosen for this key position because he is considered a rising young star. ­Having just received his MBA degree from a top American business school four years ago, he is the youngest member of upper-level management in the entire company. His business school training in the latest management science techniques has proven invaluable in improving supply chain management throughout the company. The crises still occur, but the frequent chaos of past years has been eliminated. 1 Be aware that a minimum-cost flow problem that does have capacity constraints on the arcs is sometimes referred to as a capacitated transshipment problem. We will not use this terminology. 6.2 A Case Study: The Bmz Co. Maximum Flow Problem 213 Karl has a plan for dealing with the current crisis. This will mean calling on management science once again. Background The BMZ Company is a European manufacturer of luxury automobiles. Although its cars sell well in all the developed countries, its exports to the United States are particularly important to the company. BMZ has a well-deserved reputation for providing excellent service. One key to maintaining this reputation is having a plentiful supply of automobile replacement parts readily available to the company’s numerous dealerships and authorized repair shops. These parts are mainly stored in the company’s distribution centers and then delivered promptly when needed. One of Karl Schmidt’s top priorities is avoiding shortages at these distribution centers. The company has several distribution centers in the United States. However, the closest one to the Los Angeles center is over 1,000 miles away in Seattle. Since BMZ cars are becoming especially popular in California, it is particularly important to keep the Los Angeles center well supplied. Therefore, the fact that supplies there are currently dwindling is a matter of real concern to BMZ top management—as Karl learned forcefully today. Most of the automobile replacement parts are produced at the company’s main factory in Stuttgart, Germany, along with the production of new cars. It is this factory that has been supplying the Los Angeles center with spare parts. Some of these parts are bulky, and very large numbers of certain parts are needed, so the total volume of the supplies has been relatively massive—over 300,000 cubic feet of goods arriving monthly. Now a much larger amount will be needed over the next month to replenish the dwindling inventory. The Problem The problem is to maximize the flow of automobile replacement parts from the factory in Stuttgart, Germany, to the distribution center in Los Angeles. Karl needs to execute a plan quickly for shipping as much as possible from the main factory to the distribution center in Los Angeles over the next month. He already has recognized that this is a maximum flow problem—a problem of maximizing the flow of replacement parts from the factory to this distribution center. The factory is producing far more than can be shipped along a single route to this one distribution center. Therefore, the limiting factor on how much can be shipped is the limited capacity of the company’s distribution network. This distribution network is depicted in Figure 6.6, where the nodes labeled ST and LA are the factory in Stuttgart and the distribution center in Los Angeles, respectively. There is a rail FIGURE 6.6 The BMZ Co. distribution network from its main factory in Stuttgart, ­Germany, to a distribution center in Los Angeles. RO New York 0 [8 LA Los Angeles ts i un ts uni NY [60 [40 units max.] x.] .] ax m its New Orleans [70 units max.] [50 x.] ma a m un 0 [5 ax.] ts m i n 30 u NO [ Rotterdam un its BO ma x.] [70 units max.] x.] Bordeaux ma s t i un [40 LI Lisbon ST Stuttgart 214 Chapter Six Network Optimization Problems head at the factory, so shipments first go by rail to one of three European ports: ­Rotterdam (node RO), Bordeaux (node BO), and Lisbon (node LI). They then go by ship to ports in the United States, either New York (node NY) or New Orleans (node NO). Finally, they are shipped by truck from these ports to the distribution center in Los Angeles. The organizations operating these railroads, ships, and trucks are independently owned companies that ship goods for numerous firms. Because of prior commitments to their regular customers, these companies are unable to drastically increase the allocation of space to any single customer on short notice. Therefore, the BMZ Co. is only able to secure a limited amount of shipping space along each shipping lane over the next month. The amounts available are given in Figure 6.6, using units of hundreds of cubic meters. (Since each unit of 100 cubic meters is a little over 3,500 cubic feet, these are large volumes of goods that need to be moved.) Model Formulation In contrast to the spreadsheet model in Figure 6.5, which minimizes TotalCost (D11), the spreadsheet model in Figure 6.8 ­maximizes the objective cell MaxFlow (D14). Figure 6.7 shows the network model for this maximum flow problem. Rather than showing the geographical layout of the distribution network, this network simply lines up the nodes (representing the cities) in evenly spaced columns. The arcs represent the shipping lanes, where the capacity of each arc (given in square brackets under the arc) is the amount of shipping space available along that shipping lane. The objective is to determine how much flow to send through each arc (how many units to ship through each shipping lane) to maximize the total number of units flowing from the factory in Stuttgart to the distribution center in Los Angeles. Figure 6.8 shows the corresponding spreadsheet model for this problem when using the format introduced in Figure 6.5. The main difference from the model in Figure 6.5 is the change in the objective. Since we are no longer minimizing the total cost of the flow through the network, column G in Figure 6.5 can be deleted in Figure 6.8. The objective cell MaxFlow (D14) in Figure 6.8 now needs to give the total number of units flowing from Stuttgart to Los Angeles. Thus, the equations at the bottom of the figure include D14 = I4, where I4 gives the net flow leaving Stuttgart to go to Los Angeles. As in Figure 6.5, the equations in Figure 6.8 entered into NetFlow (I4:I10) again use the difference of two SUMIF functions to calculate the net flow generated at each node. Since the objective is to maximize the flow shown in MaxFlow (D14), the Solver Parameters box specifies that this objective cell is to be ­maximized. After running Solver, the optimal solution shown in the changing cells Ship (D4:D12) is obtained for the amount that BMZ should ship through each shipping lane. FIGURE 6.7 RO [6 0] A network model for the BMZ Co. problem as a maximum flow problem, where the number in square brackets below each arc is the capacity of that arc. [50 0] [8 [4 0] ] NY BO [70] ] 0 [7 [5 0] LA [40 ] NO ] 0 [3 LI ST 6.3 Maximum Flow Problems 215 FIGURE 6.8 A spreadsheet model for the BMZ Co. maximum flow problem, including the equations entered into the objective cell MaxFlow (D14) and the other output cells NetFlow (I4:I10), as well as the other specifications needed to set up the model. The changing cells Ship (D4:D12) show the optimal shipping quantities through the distribution network obtained by Solver. A 1 B C D E F G J K H I Nodes Net Flow 50 Stuttgart 150 70 Rotterdam 0 = 0 Bordeaux 0 = 0 BMZ Co. Maximum Flow Problem 2 To From 3 Ship Capacity ≤ 4 Stuttgart Rotterdam 50 5 Stuttgart Bordeaux 70 ≤ 6 Stuttgart Lisbon 30 ≤ 40 Supply/Demand 7 Rotterdam New York 50 ≤ 60 Lisbon 0 = 0 8 Bordeaux New York 30 ≤ 40 New York 0 0 9 Bordeaux New Orleans 40 ≤ = 50 New Orleans 0 = 0 30 Los Angeles -150 10 Lisbon New Orleans 30 ≤ 11 New York Los Angeles 80 ≤ 80 12 New Orleans Los Angeles 70 ≤ 70 13 Maximum Flow 14 Solver Parameters Set Objective Cell: MaxFlow To: Max By Changing Variable Cells: Ship Subject to the Constraints: I5:I9 = SupplyDemand Ship <= Capacity Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 150 Range Name Cells Capacity From MaxFlow NetFlow Nodes Ship SupplyDemand To F4:F12 B4:B12 D14 I4:I10 H4:H10 D4:D12 K5:K9 C4:C12 I Net Flow 3 4 =SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship) 5 =SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship) 6 =SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship) 7 =SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship) 8 =SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship) 9 =SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship) 10 =SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship) 14 C D Maximum Flow =I4 However, Karl is not completely satisfied with this solution. He has an idea for doing even better. This will require formulating and solving another maximum flow problem. (This story continues in the middle of the next section.) Review Questions 6.3 1. What is the current crisis facing the BMZ Co.? 2. When formulating this problem in network terms, what is flowing through BMZ’s distribution network? From where to where? 3. What is the objective of the resulting maximum flow problem? MAXIMUM FLOW PROBLEMS Like a minimum-cost flow problem, a maximum flow problem is concerned with flow through a network. However, the objective now is different. Rather than minimizing the cost of the flow, the objective now is to find a flow plan that maximizes the amount flowing through the network. This is how Karl Schmidt was able to find a flow plan that maximizes the number of units of automobile replacement parts flowing through BMZ’s distribution network from its factory in Stuttgart to the distribution center in Los Angeles. An Application Vignette The network for transport of natural gas on the Norwegian Continental Shelf, with approximately 5,000 miles of subsea pipelines, is the world’s largest offshore pipeline network. Gassco is a company entirely owned by the Norwegian state, which operates this network. Another company that is largely state owned, StatoilHydro, is the main Norwegian supplier of natural gas to markets throughout Europe and elsewhere. Gassco and StatoilHydro together use management science techniques to optimize both the configuration of the network and the routing of the natural gas. The main model used for this routing is a multicommodity network-flow model in which the different hydrocarbons and contaminants in natural gas constitute the commodities. The objective function for the model is to maximize the total flow of the natural gas from the supply points (the offshore drilling platforms) to the demand points (typically import terminals). However, in addition to the usual supply-anddemand constraints, the model also includes constraints involving pressure-flow relationships, maximum delivery pressures, and technical pressure bounds on pipelines. Therefore, this model is a generalization of the model for the maximum flow problem described in this section. This key application of management science, along with a few others, has had a dramatic impact on the efficiency of the operation of this offshore pipeline network. The resulting accumulated savings were estimated to be approximately $2 billion in the period 1995–2008. Source: F. Romo, A. Tomasgard, L. Hellemo, M. Fodstad, B. H. Eidesen, and B. Pedersen, “Optimizing the Norwegian Natural Gas Production and Transport,” Interfaces 39, no. 1 (January–February 2009), pp. 46–56. (A link to this article is provided at www.mhhe.com/Hillier6e.) General Characteristics Except for the difference in objective (maximize flow versus minimize cost), the characteristics of the maximum flow problem are quite similar to those for the minimum-cost flow problem. However, there are some minor differences, as we will discuss after summarizing the assumptions. Assumptions of a Maximum Flow Problem The objective is to find a flow plan that maximizes the flow from the source to the sink. Although a maximum flow problem has only a single source and a single sink, variants with multiple sources and sinks also can be solved, as illustrated in the next subsection. 216 1. All flow through the network originates at one node, called the source, and terminates at one other node, called the sink. (The source and sink in the BMZ problem are the factory in Stuttgart and the distribution center in Los Angeles, respectively, as represented by nodes ST and LA in Figure 6.7.) 2. All the remaining nodes are transshipment nodes. (These are nodes RO, BO, LI, NY, and NO in the BMZ problem.) 3. Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. At the source, all arcs point away from the node. At the sink, all arcs point into the node. 4. The objective is to maximize the total amount of flow from the source to the sink. This amount is measured in either of two equivalent ways, namely, either the amount leaving the source or the amount entering the sink. (Cells D14 and I4 in Figure 6.8 use the amount leaving the source.) The source and sink of a maximum flow problem are analogous to the supply nodes and demand nodes of a minimum-cost flow problem. These are the only nodes in both problems that do not have conservation of flow (flow out equals flow in). Like the supply nodes, the source generates flow. Like the demand nodes, the sink absorbs flow. However, there are two differences between these nodes in a minimum-cost flow problem and the corresponding nodes in a maximum flow problem. One difference is that, whereas supply nodes have fixed supplies and demand nodes have fixed demands, the source and sink do not. The reason is that the objective is to maximize the flow leaving the source and entering the sink rather than fixing this amount. The second difference is that, whereas the number of supply nodes and the number of demand nodes in a minimum-cost flow problem may be more than one, there can be only one source and only one sink in a maximum flow problem. However, variants of maximum flow problems that have multiple sources and sinks can still be solved by Solver, as you now will see illustrated by the BMZ case study introduced in the preceding section. 6.3 Maximum Flow Problems 217 Continuing the Case Study with Multiple Supply Points and Multiple Demand Points Here is Karl Schmidt’s idea for how to improve upon the flow plan obtained at the end of ­Section 6.2 (as given in column D of Figure 6.8). The company has a second, smaller factory in Berlin, north of its Stuttgart factory, for producing automobile parts. Although this factory normally is used to help supply distribution centers in northern Europe, Canada, and the northern United States (including one in Seattle), it also is able to ship to the distribution center in Los Angeles. Furthermore, the distribution center in Seattle has the capability of supplying parts to the customers of the distribution center in Los Angeles when shortages occur at the latter center. In this light, Karl now has developed a better plan for addressing the current inventory shortages in Los Angeles. Rather than simply maximizing shipments from the Stuttgart factory to Los Angeles, he has decided to maximize the total shipments from both factories to the distribution centers in both Los Angeles and Seattle. Figure 6.9 shows the network model representing the expanded distribution network that encompasses both factories and both distribution centers. In addition to the nodes shown in Figures 6.6 and 6.7, node BE is the second, smaller factory in Berlin; nodes HA and BN are additional ports used by this factory in Hamburg and Boston, respectively; and node SE is the distribution center in Seattle. As before, the arcs represent the shipping lanes, where the number in square brackets below each arc is the capacity of that arc, that is, the maximum number of units that can be shipped through that shipping lane over the next month. The corresponding spreadsheet model is displayed in Figure 6.10. The format is the same as in Figure 6.8. However, the objective cell MaxFlow (D21) now gives the total flow from Stuttgart and Berlin, so D21 = I4 + I5 (as shown by the equation for this objective cell given at the bottom of the figure). The changing cells Ship (D4:D19) in this figure show the optimal solution obtained for the number of units to ship through each shipping lane over the next month. Comparing this solution with the one in Figure 6.8 shows the impact of Karl Schmidt’s decision to expand the distribution network to include the second factory and the distribution center in Seattle. As indicated in column I of the two figures, the number of units going to Los Angeles directly has been increased from 150 to 160, in addition to the 60 units going to Seattle as a backup for the inventory shortage in Los Angeles. This plan solved the crisis in Los Angeles and won Karl commendations from top management. FIGURE 6.9 HA BN ] [40 [6 0] ] [30 ] [20 SE RO [40 ] ] 0] 0] [5 ] LA BE [20] [60 NY [1 A network model for the expanded BMZ Co. problem as a variant of a maximum flow problem, where the number in square brackets below each arc is the capacity of that arc. [40 [80 [70 ] ] BO ST [70] ] [50 0] NO [4 [30 ] LI 218 Chapter Six Network Optimization Problems FIGURE 6.10 A spreadsheet model for the expanded BMZ Co. problem as a variant of a maximum flow problem with sources in both Stuttgart and Berlin and sinks in both Los Angeles and Seattle. Using the objective cell MaxFlow (D21) to maximize the total flow from the two sources to the two sinks, Solver yields the optimal shipping plan shown in the changing cells Ship (D4:D19). A 1 B C D E F G J H I Capacity Nodes Net Flow K BMZ Co. Expanded Maximum Flow Problem 2 Supply/Demand 3 From To Ship 4 Stuttgart Rotterdam 40 ≤ 50 Stuttgart 140 5 Stuttgart Bordeaux 70 ≤ 70 Berlin 80 6 Stuttgart Lisbon 30 ≤ 40 Hamburg 0 = 0 7 Berlin Rotterdam 20 ≤ 20 Rotterdam 0 = 0 8 Berlin Hamburg 60 ≤ 60 Bordeaux 0 = 0 9 Rotterdam New York 60 ≤ 60 Lisbon 0 = 0 ≤ 40 Boston 0 = 0 10 Bordeaux New York 30 11 Bordeaux New Orleans 40 ≤ 50 New York 0 = 0 12 Lisbon New Orleans 30 ≤ 30 New Orleans 0 = 0 13 Hamburg New York 30 ≤ 30 Los Angeles -160 14 Hamburg Boston 30 ≤ 40 Seattle -60 15 New Orleans Los Angeles 70 ≤ 70 16 New York Los Angeles 80 ≤ 80 17 New York Seattle 40 ≤ 40 18 Boston Los Angeles 10 ≤ 10 20 ≤ 20 19 Boston Seattle 20 21 Maximum Flow Solver Parameters Set Objective Cell: MaxFlow To: Max By Changing Variable Cells: Ship Subject to the Constraints: I6:I12 = SupplyDemand Ship <= Capacity Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 220 Range Name Cells Capacity From MaxFlow NetFlow Nodes Ship SupplyDemand To F4:F19 B4:B19 D21 I4:I14 H4:H14 D4:D19 K6:K12 C4:C19 I 3 Net Flow 4 =SUMIF(From,H4,Ship)-SUMIF(To,H4,Ship) 5 =SUMIF(From,H5,Ship)-SUMIF(To,H5,Ship) 6 =SUMIF(From,H6,Ship)-SUMIF(To,H6,Ship) 7 =SUMIF(From,H7,Ship)-SUMIF(To,H7,Ship) 8 =SUMIF(From,H8,Ship)-SUMIF(To,H8,Ship) 9 =SUMIF(From,H9,Ship)-SUMIF(To,H9,Ship) 10 =SUMIF(From,H10,Ship)-SUMIF(To,H10,Ship) 11 =SUMIF(From,H11,Ship)-SUMIF(To,H11,Ship) 12 =SUMIF(From,H12,Ship)-SUMIF(To,H12,Ship) 13 =SUMIF(From,H13,Ship)-SUMIF(To,H13,Ship) 14 =SUMIF(From,H14,Ship)-SUMIF(To,H14,Ship) 21 C D Maximum Flow =I4+I5 Some Applications The applications of maximum flow problems and their variants are somewhat similar to those for minimum-cost flow problems described in the preceding section when management’s objective is to maximize flow rather than to minimize cost. Here are some typical kinds of applications. 1. Maximize the flow through a distribution network, as for the BMZ Co. problem. 2. Maximize the flow through a company’s supply network from its vendors to its processing facilities. 6.4 Shortest Path Problems 219 3. Maximize the flow of oil through a system of pipelines. 4. Maximize the flow of water through a system of aqueducts. 5. Maximize the flow of vehicles through a transportation network. Solving Very Large Problems The expanded BMZ network in Figure 6.9 has 11 nodes and 16 arcs. However, the networks for most real applications are considerably larger, and occasionally vastly larger with many thousands of nodes and arcs. For huge networks, Solver may not be capable of solving such large maximum flow problems even if they still can be formulated as illustrated in Figures 6.8 and 6.10 . Fortunately, management scientists have other ways of efficiently formulating huge maximum flow problems and then solving them. For example, an extremely efficient ­special-purpose algorithm has been developed specifically for solving maximum flow problems. Another solution approach is to reformulate the problem to fit the format for a minimum-cost flow problem so that the network simplex method can be applied. These special algorithms are available in some software packages, but not in Solver. Thus, if you should ever encounter a maximum flow problem or a variant that is beyond the scope of Solver (which won’t happen in this book), rest assured that it probably can be formulated and solved in another way. Review Questions 6.4 1. How does the objective of a maximum flow problem differ from that for a minimum-cost flow problem? 2. What are the source and the sink for a maximum flow problem? For each, in what direction do all their arcs point? 3. What are the two equivalent ways in which the total amount of flow from the source to the sink can be measured? 4. The source and sink of a maximum flow problem are different from the supply nodes and demand nodes of a minimum-cost flow problem in what two ways? 5. What are a few typical kinds of applications of maximum flow problems? SHORTEST PATH PROBLEMS The most common applications of shortest path problems are for what the name suggests— finding the shortest path between two points. Here is an example. An Example: The Littletown Fire Department Problem The objective is to find the shortest route from the fire station to the farming community. Littletown is a small town in a rural area. Its fire department serves a relatively large geographical area that includes many farming communities. Since there are numerous roads throughout the area, many possible routes may be available for traveling to any given farming community from the fire station. Since time is of the essence in reaching a fire, the fire chief wishes to determine in advance the shortest path from the fire station to each of the farming communities. Figure 6.11 shows the road system connecting the fire station to one of the farming communities, including the mileage along each road. Can you find which route from the fire station to the farming community minimizes the total number of miles? Model Formulation for the Littletown Problem Figure 6.12 gives a network representation of this problem, which ignores the geographical ­layout and the curves in the roads. (When lining up the nodes in columns as in this figure, the selection of which nodes should go together in a column is somewhat arbitrary, but the ­selections here fit both the geography and the road system well.) This network model is the usual way of representing a shortest path problem. The junctions now are nodes of the network, where the fire station and farming community are two additional nodes labeled as O (for origin) and T An Application Vignette Incorporated in 1881, Canadian Pacific Railway (CPR) was North America’s first transcontinental railway. CPR transports rail freight over a 14,000-mile network extending from Montreal to Vancouver and throughout the U.S. Northwest and Midwest. Alliances with other carriers extend CPR’s market reach into the major business centers of Mexico as well. Every day CPR receives approximately 7,000 new shipments from its customers going to destinations across North America and for export. It must route and move these shipments in railcars over the network of track, where a railcar may be switched a number of times from one locomotive engine to another before reaching its destination. CPR must coordinate the shipments with its operational plans for approximately 1,600 locomotives, 65,000 railcars, over 5,000 train crew members, and 250 train yards. CPR management turned to a management science consulting firm, MultiModal Applied Systems, to work with CPR employees in developing a management science approach to this problem. A variety of management science techniques were used to create a new operating strategy. However, the foundation of the approach was to represent the flow of blocks links In a shortest path problem, travel goes from the origin to the destination through a series of links (such as roads) that connect pairs of nodes (junctions) in the network. of railcars as flow through a network where each node corresponds to both a location and a point in time. This representation then enabled the application of network optimization techniques. For example, numerous shortest path problems are solved each day as part of the overall approach. This application of management science is saving CPR roughly US$100 million per year. Labor productivity, locomotive productivity, fuel consumption, and railcar velocity have improved very substantially. In addition, CPR now provides its customers with reliable delivery times and has received many awards for its improvement in service. This application of network optimization techniques also led to CPR winning the prestigious First Prize in the 2003 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences. Source: P. Ireland, R. Case, J. Fallis, C. Van Dyke, J. Kuehn, and M. Meketon, “The Canadian Pacific Railway Transforms Operations by Using Models to Develop Its Operating Plans,” Interfaces 34, no. 1 (January–February 2004), pp. 5–14. (A link to this article is provided at www.mhhe.com/Hillier6e.) (for destination), respectively. Since travel (flow) can go in either direction between the nodes, the lines connecting the nodes now are referred to as links2 instead of arcs. A link between a pair of nodes allows travel in either direction, whereas an arc allows travel in only the direction indicated by an arrowhead, so the lines in Figure 6.12 need to be links instead of arcs. (Notice that the links do not have an arrowhead at either end.) 8 FIGURE 6.11 The road system between the Littletown Fire Station and a certain farming community, where A, B, . . ., H are junctions and the number next to each road shows its distance in miles. 6 A 1 3 6 Fire Station D 4 6 3 (Origin) O 1 6 4 2 C 2 220 6 5 7 3 H 4 3 6 5 E 2 F 8 D 4 B 4 7 A The network representation of Figure 6.11 as a shortest path problem. 5 E 2 C Farming Community G 5 FIGURE 6.12 6 3 3 B 4 4 F 4 G 2 6 T (Destination) 7 H Another name sometimes used is undirected arc, but we will not use this terminology. 7 6.4 Shortest Path Problems 221 Have you found the shortest path from the origin to the destination yet? (Try it now before reading further.) It is O → A → B → E → F → T This spreadsheet model is like one for a minimum-cost flow problem with no arc capacity constraints except that distances replace unit costs and travel on a chosen path is interpreted as a flow of 1 through this path. with a total distance of 19 miles. This problem (like any shortest path problem) can be thought of as a special kind of minimum-cost flow problem (Section 6.1) where the miles traveled now are interpreted to be the cost of flow through the network. A trip from the fire station to the farming community is interpreted to be a flow of 1 on the chosen path through the network, so minimizing the cost of this flow is equivalent to minimizing the number of miles traveled. The fire station is considered to be the one supply node, with a supply of 1 to represent the start of this trip. The farming community is the one demand node, with a demand of 1 to represent the completion of this trip. All the other nodes in Figure 6.12 are transshipment nodes, so the net flow generated at each is 0. Figure 6.13 shows the spreadsheet model that results from this interpretation. The format is basically the same as for the minimum-cost flow problem formulated in Figure 6.5, except now there are no arc capacity constraints and the unit cost column is replaced by a column of distances in miles. The flow quantities given by the changing cells OnRoute (D4:D27) are 1 for each arc that is on the chosen path from the fire station to the farming community and 0 otherwise. The objective cell TotalDistance (D29) gives the total distance of this path in miles. (See the equation for this cell at the bottom of the figure.) Columns B and C together list all the vertical links in Figure 6.12 twice, once as a downward arc and once as an upward arc, since either direction might be on the chosen path. The other links are only listed as leftto-right arcs, since this is the only direction of interest for choosing a shortest path from the origin to the destination. Column K shows the net flow that needs to be generated at each of the nodes. Using the equations at the bottom of the figure, each column I cell then calculates the actual net flow at that node by adding the flow out and subtracting the flow in. The corresponding constraints, Nodes (H4:H13) = SupplyDemand (K4:K13), are specified in the Solver Parameters box. The solution shown in OnRoute (D4:D27) is the optimal solution obtained after running Solver. It is exactly the same as the shortest path given earlier. Just as for minimum-cost flow problems and maximum flow problems, special algorithms are available for solving large shortest path problems very efficiently, but these algorithms are not included in Solver. Using a spreadsheet formulation and Solver is fine for problems of the size of the Littletown problem and somewhat larger, but you should be aware that vastly larger problems can still be solved by other means. General Characteristics Except for more complicated variations beyond the scope of this book, all shortest path problems share the characteristics illustrated by the Littletown problem. Here are the basic assumptions. Assumptions of a Shortest Path Problem The objective is to find the shortest path from the ­origin to the destination. 1. You need to choose a path through the network that starts at a certain node, called the ­origin, and ends at another certain node, called the destination. 2. The lines connecting certain pairs of nodes commonly are links (which allow travel in either direction), although arcs (which only permit travel in one direction) also are allowed. 3. Associated with each link (or arc) is a nonnegative number called its length. (Be aware that the drawing of each link in the network typically makes no effort to show its true length other than giving the correct number next to the link.) 4. The objective is to find the shortest path (the path with the minimum total length) from the origin to the destination. Some Applications Not all applications of shortest path problems involve minimizing the distance traveled from the origin to the destination. In fact, they might not even involve travel at all. The links (or arcs) might instead represent activities of some other kind, so choosing a path through the 222 Chapter Six Network Optimization Problems FIGURE 6.13 A spreadsheet model for the Littletown Fire Department shortest path problem, including the equations entered into the objective cell TotalDistance (D29) and the other output cells SupplyDemand (K4:K13). The values of 1 in the changing cells OnRoute (D4:D27) reveal the optimal solution obtained by Solver for the shortest path (19 miles) from the fire station to the farming community. A 1 B C D E F G H I J K Littletown Fire Department Shortest Path Problem 2 3 From To On Route Distance Nodes Net Flow 4 Supply/Demand Fire St. A 1 3 Fire St. 1 = 1 5 Fire St. B 0 6 A 0 = 0 6 Fire St. C 0 4 B 0 = 0 7 A B 1 1 C 0 = 0 8 A D 0 6 D 0 = 0 9 B A 0 1 E 0 = 0 10 B C 0 2 F 0 = 0 11 B D 0 4 G 0 = 0 12 B E 1 5 H 0 = 0 13 C B 0 2 Farm Comm. –1 = –1 14 C E 0 7 15 D E 0 3 16 D F 0 8 17 E D 0 3 18 E F 1 6 19 E G 0 5 20 E H 0 4 21 F G 0 3 22 F Farm Comm. 1 4 23 G F 0 3 24 G H 0 2 25 G Farm Comm. 0 6 26 H G 0 2 27 H Farm Comm. 0 7 Total Distance 19 28 29 Solver Parameters I Set Objective Cell: TotalDistance To: Min By Changing Variable Cells: OnRoute Subject to the Constraints: NetFlow = SupplyDemand Solver Options: Make Variables Nonnegative Solving Method: Simplex LP Range Name Cells Distance From NetFlow Nodes OnRoute SupplyDemand To TotalDistance F4:F27 B4:B27 I4:I13 H4:H13 D4:D27 K4:K13 C4:C27 D29 3 Net Flow 4 =SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute) 5 =SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute) 6 =SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute) 7 =SUMIF(From,H7,OnRoute)-SUMIF(To,H7,OnRoute) 8 =SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute) 9 =SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute) 10 =SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute) 11 =SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute) 12 =SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute) 13 =SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute) 29 C D Total Distance =SUMPRODUCT(OnRoute,Distance) 6.4 Shortest Path Problems 223 network corresponds to selecting the best sequence of activities. The numbers giving the “lengths” of the links might then be, for example, the costs of the activities, in which case the objective would be to determine which sequence of activities minimizes the total cost. Here are three categories of applications. 1. Minimize the total distance traveled, as in the Littletown example. 2. Minimize the total cost of a sequence of activities, as in the example that follows in the subsection below. 3. Minimize the total time of a sequence of activities, as in the example involving the Quick Company at the end of this section. An Example of Minimizing Total Cost Sarah needs a schedule for trading in her car that will minimize her total net cost. Sarah has just graduated from high school. As a graduation present, her parents have given her a car fund of $21,000 to help purchase and maintain a certain three-year-old used car for college. Since operating and maintenance costs go up rapidly as the car ages, Sarah’s parents tell her that she will be welcome to trade in her car on another three-year-old car one or more times during the next three summers if she determines that this would minimize her total net cost. They also inform her that they will give her a new car in four years as a college graduation present, so she should definitely plan to trade in her car then. (These are pretty nice parents!) Table 6.2 gives the relevant data for each time Sarah purchases a three-year-old car. For example, if she trades in her car after two years, the next car will be in ownership year 1 during her junior year, and so forth. When should Sarah trade in her car (if at all) during the next three summers to minimize her total net cost of purchasing, operating, and maintaining the car(s) over her four years of college? Figure 6.14 shows the network formulation of this problem as a shortest path problem. Nodes 1, 2, 3, and 4 are the end of Sarah’s first, second, third, and fourth years of college, respectively. Node 0 is now, before starting college. Each arc from one node to a second node corresponds to the activity of purchasing a car at the time indicated by the first of these two nodes and then trading it in at the time indicated by the second node. Sarah begins by purchasing a car now, and she ends by trading in a car at the end of year 4, so node 0 is the origin and node 4 is the destination. The number of arcs on the path chosen from the origin to the destination indicates how many times Sarah will purchase and trade in a car. For example, consider the path 0 TABLE 6.2 Sarah’s Data Each Time She Purchases a ThreeYear-Old Car 1 4 Operating and Maintenance Costs for Ownership Year Purchase Price $12,000 Trade-in Value at End of Ownership Year 1 2 3 4 1 2 3 4 $2,000 $3,000 $4,500 $6,500 $8,500 $6,500 $4,500 $3,000 25,000 FIGURE 6.14 Formulation of the problem of when Sarah should trade in her car as a shortest path problem. The node labels measure the number of years from now. Each arc represents purchasing a car and then trading it in later. 3 17,000 10,500 10,500 (Origin) 0 5,500 1 5,500 2 5,500 10,500 17,000 3 5,500 4 (Destination) 224 Chapter Six Network Optimization Problems This corresponds to purchasing a car now, then trading it in at the end of year 1 to purchase a second car, then trading in the second car at the end of year 3 to purchase a third car, and then trading in this third car at the end of year 4. Since Sarah wants to minimize her total net cost from now (node 0) to the end of year 4 (node 4), each arc length needs to measure the net cost of that arc’s cycle of purchasing, maintaining, and trading in a car. Therefore, Arc length = Purchase price + Operating and maintenance costs − Trade-in value For example, consider the arc from node 1 to node 3. This arc corresponds to purchasing a car at the end of year 1, operating and maintaining it during ownership years 1 and 2, and then trading it in at the end of ownership year 2. Consequently, Length of arc from ① to ③ = 12,000 + 2,000 + 3,000 − 6,500 = 10,500 (in dollars) The sum of the arc lengths on any path through this network gives the total net cost of the corresponding plan for trading in cars. The objective cell now is TotalCost instead of TotalDistance. The arc lengths calculated in this way are shown next to the arcs in Figure 6.14. Adding up the lengths of the arcs on any path from node 0 to node 4 then gives the total net cost for that particular plan for trading in cars over the next four years. Therefore, finding the shortest path from the origin to the destination identifies the plan that will minimize Sarah’s total net cost. Figure 6.15 shows the corresponding spreadsheet model, formulated in just the same way as for Figure 6.13 except that distances are now costs. Thus, the objective cell TotalCost (D23) now gives the total cost that is to be minimized. The changing cells OnRoute (D12:D21) in the figure display the optimal solution obtained after running Solver. Since values of 1 indicate the path being followed, the shortest path turns out to be 0 2 4 Trade in the first car at the end of year 2. Trade in the second car at the end of year 4. The length of this path is 10,500 + 10,500 = 21,000, so Sarah’s total net cost is $21,000, as given by the objective cell. Recall that this is exactly the amount in Sarah’s car fund provided by her parents. (These are really nice parents!) An Example of Minimizing Total Time The objective is to minimize the total time for the project. The Quick Company has learned that a competitor is planning to come out with a new kind of product with great sales potential. Quick has been working on a similar product that had been scheduled to come to market in 20 months. However, research is nearly complete and Quick’s management now wishes to rush the product out to meet the competition. There are four nonoverlapping phases left to be accomplished, including the remaining research (the first phase) that currently is being conducted at a normal pace. However, each phase can instead be conducted at a priority or crash level to expedite completion. These are the only levels that will be considered for the last three phases, whereas both the normal level and these two levels will be considered for the first phase. The times required at these levels are shown in Table 6.3. Management now has allocated $30 million for these four phases. The cost of each phase at the levels under consideration is shown in Table 6.4. Management wishes to determine at which level to conduct each of the four phases to minimize the total time until the product can be marketed, subject to the budget restriction of $30 million. Figure 6.16 shows the network formulation of this problem as a shortest path problem. Each node indicates the situation at that point in time. Except for the destination, a node is identified by two numbers: 1. The number of phases completed. 2. The number of millions of dollars left for the remaining phases. 6.4 Shortest Path Problems 225 FIGURE 6.15 A spreadsheet model that formulates Sarah’s problem as a shortest path problem where the objective is to minimize the total cost instead of the total distance. The bottom of the figure shows the equations entered in the objective cell TotalCost (D23) and the other output cells Cost (E12:E21) and NetFlow (H12:H16). After applying Solver, the values of 1 in the changing cells OnRoute (D12:D21) identify the shortest (least expensive) path for scheduling trade-ins. A 1 B C D E F G H I J Sarah's Car Purchasing Problem 2 Trade-in Operating & Value at End Purchase Maint. Cost of Year Price 3 4 5 Year 1 $2,000 $8,500 6 Year 2 $3,000 $6,500 7 Year 3 $4,500 $4,500 8 Year 4 $6,500 $3,000 $12,000 9 10 11 From To On Route Cost Nodes Net Flow 12 Year 0 Year 1 0 $5,500 Year 0 1 = Supply/Demand 1 13 Year 0 Year 2 1 $10,500 Year 1 0 = 0 14 Year 0 Year 3 0 $17,000 Year 2 0 = 0 15 Year 0 Year 4 0 $25,000 Year 3 0 = 0 16 Year 1 Year 2 0 $5,500 Year 4 -1 = -1 17 Year 1 Year 3 0 $10,500 18 Year 1 Year 4 0 $17,000 19 Year 2 Year 3 0 $5,500 20 Year 2 Year 4 1 $10,500 21 Year 3 Year 4 0 $5,500 Total Cost $21,000 22 23 Range Name Cells Cost From NetFlow Nodes OnRoute OpMaint1 OpMaint2 OpMaint3 OpMaint4 PurchasePrice SupplyDemand To TotalCost TradeIn1 TradeIn2 TradeIn3 TradeIn4 E12:E21 B12:B21 H12:H16 G12:G16 D12:D21 C5 C6 C7 C8 E5 J12:J16 C12:C21 D23 D5 D6 D7 D8 E Cost 11 12 =PurchasePrice+OpMaint1-TradeIn1 13 =PurchasePrice+OpMaint1+OpMaint2-TradeIn2 14 =PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3 15 =PurchasePrice+OpMaint1+OpMaint2+OpMaint3+OpMaint4-TradeIn4 16 =PurchasePrice+OpMaint1-TradeIn1 17 =PurchasePrice+OpMaint1+OpMaint2-TradeIn2 18 =PurchasePrice+OpMaint1+OpMaint2+OpMaint3-TradeIn3 19 =PurchasePrice+OpMaint1-TradeIn1 20 =PurchasePrice+OpMaint1+OpMaint2-TradeIn2 21 =PurchasePrice+OpMaint1-TradeIn1 H Net Flow 11 Solver Parameters 12 =SUMIF(From,G12,OnRoute)-SUMIF(To,G12,OnRoute) Set Objective Cell: TotalCost To: Min By Changing Variable Cells: OnRoute Subject to the Constraints: NetFlow = SupplyDemand 13 =SUMIF(From,G13,OnRoute)-SUMIF(To,G13,OnRoute) 14 =SUMIF(From,G14,OnRoute)-SUMIF(To,G14,OnRoute) 15 =SUMIF(From,G15,OnRoute)-SUMIF(To,G15,OnRoute) 16 =SUMIF(From,G16,OnRoute)-SUMIF(To,G16,OnRoute) Solver Options: Make Variables Nonnegative Solving Method: Simplex LP C 23 Total Cost D =SUMPRODUCT(OnRoute,Cost) The origin is now, when 0 phases have been completed and the entire budget of $30 million is left. Each arc represents the choice of a particular level of effort (identified in parentheses below the arc) for that phase. [There are no crash arcs emanating from the (2, 12) and (3, 3) nodes because this level of effort would require exceeding the budget of $30 million for the four phases.] The time (in months) required to perform the phase with this level of effort 226 Chapter Six Network Optimization Problems TABLE 6.3 Time Required for the Phases of Preparing Quick Co.’s New Product Level Remaining Research Development Design of Manufacturing System Initiate Production and Distribution Normal Priority Crash 5 months 4 months 2 months — 3 months 2 months — 5 months 3 months — 2 months 1 month Level Remaining Research Development Design of Manufacturing System Initiate Production and Distribution Normal Priority Crash $3 million 6 million 9 million — $6 million 9 million — $ 9 million 12 million — $3 million 6 million TABLE 6.4 Cost for the Phases of Preparing Quick Co.’s New Product FIGURE 6.16 Formulation of the Quick Co. problem as a shortest path problem. Except for the dummy destination, the arc labels indicate, first, the number of phases completed and, second, the amount of money left (in millions of dollars) for the remaining phases. Each arc length gives the time (in months) to perform that phase. The sum of the arc lengths on any path through this network gives the total time of the corresponding plan for preparing the new product. 5 3 2, 21 (Priority) y) t i r o (Pri (C 3 ra 1, 27 sh 2 ) ( 5 C ) l 5 r a a s h ) m 2, 18 or (Priority) 3 (Origin) N ( ty) 4 iori (C 3 0, 30 1, 24 (Pr ra 2 sh (Priority) ) (Cra s 2 5 h ) (C 2, 15 ra 3 (Priority) sh ) ity) rior (C 3 1, 21 (P ra 2 sh ) (Cra sh) 5 2, 12 (Priority) 2 (Priority) (C 1 ra sh ) 2 3, 9 (Priority) (C 1 ra sh ) 2 3, 6 (Priority) (C 1 ra sh ) 2 3, 3 (Priority) 3, 12 4, 9 0 4, 6 0 (Destination) T 0 4, 3 0 4, 0 then is the length of the arc (shown above the arc). Time is chosen as the measure of arc length because the objective is to minimize the total time for all four phases. Summing the arc lengths for any particular path through the network gives the total time for the plan corresponding to that path. Therefore, the shortest path through the network identifies the plan that minimizes total time. All four phases have been completed as soon as any one of the four nodes with a first label of 4 has been reached. So why doesn’t the network just end with these four nodes rather than having an arc coming out of each one? The reason is that a shortest path problem is required to have only a single destination. Consequently, a dummy destination is added at the righthand side. When real travel through a network can end at more than one node, an arc with length 0 is inserted from each of these nodes to a dummy destination so that the network will have just a single destination. The objective cell now is TotalTime instead of TotalDistance. Since each of the arcs into the dummy destination has length 0, this addition to the network does not affect the total length of a path from the origin to its ending point. Figure 6.17 displays the spreadsheet model for this problem. Once again, the format is the same as in Figures 6.13 and 6.15, except now the quantity of concern in column F and the objective cell TotalTime (D32) is time rather than distance or cost. Since Solver has already been run, the changing cells OnRoute (D4:D30) indicate which arcs lie on the path that minimizes the total time. Thus, the shortest path is 0, 30 1, 21 2, 15 3, 3 4, 0 T 6.4 Shortest Path Problems 227 with a total length of 2 + 3 + 3 + 2 + 0 = 10 months, as given by TotalTime (D32). The resulting plan for the four phases is shown in Table 6.5. Although this plan does consume the entire budget of $30 million, it reduces the time until the product can be brought to market from the originally planned 20 months down to just 10 months. Given this information, Quick’s management now must decide whether this plan provides the best trade-off between time and cost. What would be the effect on total time of spending a few million more dollars? What would be the effect of reducing the spending somewhat instead? It is easy to provide management with this information as well by quickly solving FIGURE 6.17 A spreadsheet model that formulates the Quick Co. problem as a shortest path problem where the objective is to minimize the total time instead of the total distance, so the objective cell is TotalTime (D32). The other output cells are NetFlow (I4:I20). The values of 1 in the changing cells OnRoute (D4:D30) reveal the shortest (quickest) path obtained by Solver. A 1 B C D E F G H I J K Quick Co. Product Development Scheduling Problem 2 3 From To On Route Time Nodes Net Flow 4 (0, 30) (1, 27) 0 5 (0, 30) 1 = 1 5 (0, 30) (1, 24) 0 4 (1, 27) 0 = 0 6 (0, 30) (1, 21) 1 2 (1, 24) 0 = 0 7 (1, 27) (2, 21) 0 3 (1, 21) 0 = 0 8 (1, 27) (2, 18) 0 2 (2, 21) 0 = 0 Supply/Demand 9 (1, 24) (2, 18) 0 3 (2, 18) 0 = 0 10 (1, 24) (2, 15) 0 2 (2, 15) 0 = 0 11 (1, 21) (2, 15) 1 3 (2, 12) 0 = 0 12 (1, 21) (2, 12) 0 2 (3, 12) 0 = 0 13 (2, 21) (3, 12) 0 5 (3, 9) 0 = 0 14 (2, 21) (3, 9) 0 3 (3, 6) 0 = 0 15 (2, 18) (3, 9) 0 5 (3, 3) 0 = 0 16 (2, 18) (3, 6) 0 3 (4, 9) 0 = 0 17 (2, 15) (3, 6) 0 5 (4, 6) 0 = 0 18 (2, 15) (3, 3) 1 3 (4, 3) 0 = 0 19 (2, 12) (3, 3) 0 5 (4, 0) 0 = 0 20 (3, 12) (4, 9) 0 2 (T) -1 = -1 21 (3, 12) (4, 6) 0 1 22 (3, 9) (4, 6) 0 2 23 (3, 9) (4, 3) 0 1 24 (3, 6) (4, 3) 0 2 25 (3, 6) (4, 0) 0 1 26 (3, 3) (4, 0) 1 2 27 (4, 9) (T) 0 0 28 (4, 6) (T) 0 0 29 (4, 3) (T) 0 0 (4, 0) (T) 1 0 30 31 32 Total Time 10 (continued) 228 Chapter Six Network Optimization Problems FIGURE 6.17 (continued) Range Name Cells From NetFlow Nodes OnRoute SupplyDemand Time To TotalTime B4:B30 I4:I20 H4:H20 D4:D30 K4:K20 F4:F30 C4:C30 D32 Solver Parameters Set Objective Cell: TotalTime To: Min By Changing Variable Cells: OnRoute Subject to the Constraints: NetFlow = SupplyDemand Solver Options: Make Variables Nonnegative Solving Method: Simplex LP I Net Flow 3 4 =SUMIF(From,H4,OnRoute)-SUMIF(To,H4,OnRoute) 5 =SUMIF(From,H5,OnRoute)-SUMIF(To,H5,OnRoute) 6 =SUMIF(From,H6,OnRoute)-SUMIF(To,H6,OnRoute) 7 =SUMIF(From,H7, OnRoute)-SUMIF(To,H7,OnRoute) 8 =SUMIF(From,H8,OnRoute)-SUMIF(To,H8,OnRoute) 9 =SUMIF(From,H9,OnRoute)-SUMIF(To,H9,OnRoute) 10 =SUMIF(From,H10,OnRoute)-SUMIF(To,H10,OnRoute) 11 =SUMIF(From,H11,OnRoute)-SUMIF(To,H11,OnRoute) 12 =SUMIF(From,H12,OnRoute)-SUMIF(To,H12,OnRoute) 13 =SUMIF(From,H13,OnRoute)-SUMIF(To,H13,OnRoute) 14 =SUMIF(From,H14,OnRoute)-SUMIF(To,H14,OnRoute) 15 =SUMIF(From,H15,OnRoute)-SUMIF(To,H15,OnRoute) 16 =SUMIF(From,H16,OnRoute)-SUMIF(To,H16,OnRoute) 17 =SUMIF(From,H17,OnRoute)-SUMIF(To,H17,OnRoute) 18 =SUMIF(From,H18,OnRoute)-SUMIF(To,H18,OnRoute) 19 =SUMIF(From,H19,OnRoute)-SUMIF(To,H19,OnRoute) 20 =SUMIF(From,H20,OnRoute)-SUMIF(To,H20,OnRoute) 32 C D Total Time =SUMPRODUCT(OnRoute,Time) some shortest path problems that correspond to budgets different from $30 million. The ultimate decision regarding which plan provides the best time–cost trade-off then is a judgment decision that only management can make. TABLE 6.5 The Optimal Solution Obtained by Solver for Quick Co.’s Shortest Path Problem Phase Level Time Cost Remaining research Development Design of manufacturing system Initiate production and distribution Crash Priority Crash Priority 2 months 3 months 3 months 2 months $ 9 million 6 million 12 million 3 million 10 months $30 million Total Review Questions 1. What are the origin and the destination in the Littletown Fire Department example? 2. What is the distinction between an arc and a link? 3. What are the supply node and the demand node when a shortest path problem is interpreted as a minimum-cost flow problem? With what supply and demand? 4. What are three measures of the length of a link (or arc) that lead to three categories of applications of shortest path problems? 5. What is the objective for Sarah’s shortest path problem? 6. When does a dummy destination need to be added to the formulation of a shortest path problem? 7. What kind of trade-off does the management of the Quick Co. need to consider in making its final decision about how to expedite its new product to market? Chapter 6 Glossary 229 6.5 Summary Networks of some type arise in a wide variety of contexts. Network representations are very useful for portraying the relationships and connections between the components of systems. Each component is represented by a point in the network called a node, and then the connections between components (nodes) are represented by lines called arcs (for one-way travel) or links (for two-way travel). Frequently, a flow of some type must be sent through a network, so a decision needs to be made about the best way to do this. The kinds of network optimization models introduced in this chapter provide a powerful tool for making such decisions. The model for minimum-cost flow problems plays a central role among these network optimization models, both because it is so broadly applicable and because it can be readily solved. Solver solves spreadsheet formulations of reasonable size, and the network simplex method can be used to solve larger problems, including huge problems with many thousands of nodes and arcs. A minimum-cost flow problem typically is concerned with optimizing the flow of goods through a network from their points of origin (the supply nodes) to where they are needed (the demand nodes). The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. One typical application (among several) is to optimize the operation of a distribution network. Special types of minimum-cost flow problems include transportation problems and assignment problems (discussed in Chapter 3) as well as two prominent types introduced in this chapter: maximum flow problems and shortest path problems. Given the limited capacities of the arcs in the network, the objective of a maximum flow problem is to maximize the total amount of flow from a particular point of origin (the source) to a particular terminal point (the sink). For example, this might involve maximizing the flow of goods through a company’s supply network from its vendors to its processing facilities. A shortest path problem also has a beginning point (the origin) and an ending point (the destination), but now the objective is to find a path from the origin to the destination that has the minimum total length. For some applications, length refers to distance, so the objective is to minimize the total distance traveled. However, some applications instead involve minimizing either the total cost or the total time of a sequence of activities. Glossary arc A channel through which flow may occur from one node to another, shown as an arrow between the nodes pointing in the direction in which flow is allowed. (Section 6.1), 207 assignment problem A special type of minimum-cost flow problem that previously was described in Section 3.6. (Section 6.1), 211 capacity of an arc The maximum amount of flow allowed through the arc. (Section 6.1), 207 conservation of flow Having the amount of flow out of a node equal the amount of flow into that node. (Section 6.1), 207 demand node A node where the net amount of flow generated (outflow minus inflow) is a fixed negative number, so that flow is absorbed there. (Section 6.1), 207 destination The node at which travel through the network is assumed to end for a shortest path problem. (Section 6.4), 221 dummy destination A fictitious destination introduced into the formulation of a shortest path problem with multiple possible termination points to satisfy the requirement that there be just a single destination. (Section 6.4), 226 length of a link or arc The number (typically a distance, a cost, or a time) associated with including the link or arc in the selected path for a shortest path problem. (Section 6.4), 221 link A channel through which flow may occur in either direction between a pair of nodes, shown as a line between the nodes. ­(Section 6.4), 220 network simplex method A streamlined version of the simplex method for solving ­minimum-cost flow problems very efficiently. (Section 6.1), 210 node A junction point of a network, shown as a labeled circle. (Section 6.1), 207 origin The node at which travel through the network is assumed to start for a shortest path problem. (Section 6.4), 221 sink The node for a maximum flow problem at which all flow through the network terminates. (Section 6.3), 216 source The node for a maximum flow problem at which all flow through the network originates. (Section 6.3), 216 supply node A node where the net amount of flow generated (outflow minus inflow) is a fixed positive number. (Section 6.1), 207 transportation problem A special type of minimum-cost flow problem that previously was described in Section 3.5. (Section 6.1), 211 transshipment node A node where the amount of flow out equals the amount of flow in. (Section 6.1), 207 transshipment problem A special type of minimum-cost flow problem where there are no capacity constraints on the arcs. (Section 6.1), 211 230 Chapter Six Network Optimization Problems Learning Aids for This Chapter All learning aids are available at www.mhhe.com/Hillier6e. Sarah Example Excel Files: Quick Example Distribution Unlimited Example Excel Add-In: BMZ Example Analytic Solver Expanded BMZ Example Supplement to This Chapter: Littletown Fire Department Example Minimum Spanning-Tree Problems Solved Problems The solutions are available at www.mhhe.com/Hillier6e. 6.S1. Distribution at Heart Beats Heart Beats is a manufacturer of medical equipment. The company’s primary product is a device used to monitor the heart during medical procedures. This device is produced in two factories and shipped to two warehouses. The product is then shipped on demand to four third-party wholesalers. All shipping is done by truck. The product distribution network is shown below. The annual production capacity at factories 1 and 2 is 400 and 250, respectively. The annual demand at wholesalers 1, 2, 3, and 4 is 200, 100, 150, and 200, respectively. The cost of shipping one unit in each shipping lane is shown on the arcs. Because of limited truck capacity, at most 250 units can be shipped from Factory 1 to Warehouse 1 each year. Formulate and solve a network optimization model in a spreadsheet to determine how to distribute the product at the lowest possible annual cost. WS 2 F1 $40 [250] WH 1 $35 $35 WS 2 Exxo 76 is an oil company that operates the pipeline network shown below, where each pipeline is labeled with its maximum flow rate in million cubic feet (MMcf) per day. A new oil well has been constructed near A. They would like to transport oil from the well near A to their refinery at G. Formulate and solve a network optimization model to determine the maximum flow rate from A to G. F2 $25 [–200] [–100] A 20 2 $50 WS 3 [–150] 6.S3. Driving to the Mile-High City Sarah and Jennifer have just graduated from college at the University of Washington in Seattle and want to go on a road trip. They have always wanted to see the mile-high city of Denver. Their road atlas shows the driving time (in hours) between various city pairs, as shown below. Formulate and solve a network optimization model to find the quickest route from Seattle to Denver. Portland 9 7 Butte 10 4 7 7 Boise 6 $65 WS 4 16 15 23 G 4 F 8 C 14 6 12 Seattle $55 WH 2 B 20 3 [250] D 15 E $60 [400] 6.S2. Assessing the Capacity of a Pipeline Network 12 7 14 Salt Lake 5 City [–200] Billings 7 Cheyenne 1 Grand 4 Denver Junction Chapter 6 Problems 231 Problems Problems An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 6.1. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 6.1. Briefly describe how the model for a special type of minimum-cost flow problem was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 6.2.* Consider the transportation problem having the unit shipping costs shown in the following table. Destination 1 2 3 Supply Source 1 2 6 5 7 8 4 6 Demand 30 40 30 40 60 a. Formulate a network model for this problem as a minimum-cost flow problem by drawing a network similar to Figure 6.3. b. Formulate and solve a spreadsheet model for this problem. 6.3. Reconsider Problem 6.2. Suppose now that the demand at each of the destinations has been reduced by 10, so that the new demands are 20, 30, and 20 at destinations 1, 2, and 3, respectively. Introduce a dummy destination in order to satisfy the feasible solutions property and then repeat parts a and b to determine how much each source should supply to each destination for this new version of the problem, a. Formulate and solve a spreadsheet model for this problem without introducing a dummy destination. b. Now convert this problem into a minimum-flow cost flow problem by introducing a dummy destination and making the other needed adjustments. Then draw a network similar to Figure 6.3 for this minimum-cost flow problem.) c. Formulate and solve a spreadsheet model for the minimum-cost flow problem obtained in part b. Then compare the resulting optimal flow plan (while ignoring the flows into the dummy destination) with the optimal flow plan obtained in part a without using a dummy destination. 6.4. The Makonsel Company is a fully integrated company that both produces goods and sells them at its retail outlets. After production, the goods are stored in the company’s two warehouses until needed by the retail outlets. Trucks are used to transport the goods from the two plants to the warehouses, and then from the warehouses to the three retail outlets. Using units of full truckloads, the first table below shows each plant’s monthly output, its shipping cost per truckload sent to each warehouse, and the maximum amount that it can ship per month to each warehouse. For each retail outlet (RO), the second table below shows its monthly demand, its shipping cost per truckload from each warehouse, and the maximum amount that can be shipped per month from each warehouse. Management now wants to determine a distribution plan (number of truckloads shipped per month from each plant to each warehouse and from each warehouse to each retail outlet) that will minimize the total shipping cost. a. Draw a network that depicts the company’s distribution network. Identify the supply nodes, transshipment nodes, and demand nodes in this network. b. Formulate a network model for this problem as a minimum-cost flow problem by inserting all the necessary data into the network drawn in part a. (Use the format depicted in Figure 6.3 to display these data.) c. Formulate and solve a spreadsheet model for this problem. Unit Shipping Cost Shipping Capacity To From Warehouse 1 Warehouse 2 Warehouse 1 Warehouse 2 Output Plant 1 Plant 2 $425 510 $560 600 125 175 150 200 200 300 Unit Shipping Cost Shipping Capacity To From Warehouse 1 Warehouse 2 Demand RO1 RO2 RO3 RO1 RO2 RO3 $470 390 $505 410 $490 440 100 125 150 150 100 75 150 200 150 150 200 150 232 Chapter Six Network Optimization Problems 6.5. The Audiofile Company produces boomboxes. However, management has decided to subcontract out the production of the speakers needed for the boomboxes. Three vendors are available to supply the speakers. Their price for each shipment of 1,000 speakers is shown below. Vendor Price 1 2 3 $22,500 22,700 22,300 Each shipment would go to one of the company’s two warehouses. In addition to the price for each shipment, each vendor would charge a shipping cost for which it has its own formula based on the mileage to the warehouse. These formulas and the mileage data are shown below. Vendor Charge per Shipment Warehouse 1 Warehouse 2 1 2 3 $300 + 40¢/mile $200 + 50¢/mile $500 + 20¢/mile 1,600 miles 500 miles 2,000 miles 400 miles 600 miles 1,000 miles Whenever one of the company’s two factories needs a shipment of speakers to assemble into the boomboxes, the company hires a trucker to bring the shipment in from one of the warehouses. The cost per shipment is given next, along with the number of shipments needed per month at each factory. Unit Shipping Cost Factory 1 Factory 2 Warehouse 1 $200 $700 Warehouse 2 400 500 10 6 Monthly demand Each vendor is able to supply as many as 10 shipments per month. However, because of shipping limitations, each vendor is only able to send a maximum of six shipments per month to each warehouse. Similarly, each warehouse is only able to send a maximum of six shipments per month to each factory. Management now wants to develop a plan for each month regarding how many shipments (if any) to order from each vendor, how many of those shipments should go to each warehouse, and then how many shipments each warehouse should send to each factory. The objective is to minimize the sum of the purchase costs (including the shipping charge) and the shipping costs from the warehouses to the factories. a. Draw a network that depicts the company’s supply network. Identify the supply nodes, transshipment nodes, and demand nodes in this network. b. This problem is only a variant of a minimum-cost flow problem because the supply from each vendor is a maximum of 10 rather than a fixed amount of 10. However, it can be converted to a full-fledged minimumcost flow problem by adding a dummy demand node that receives (at zero cost) all the unused supply capacity at the vendors. Formulate a network model for this minimum-cost flow problem by inserting all the necessary data into the network drawn in part a supplemented by this dummy demand node. (Use the format depicted in Figure 6.3 to display these data.) c. Formulate and solve a spreadsheet model for the company’s problem. 6.6.* Consider Figure 6.9 (in Section 6.3), which depicts the BMZ Co. distribution network from its factories in Stuttgart and Berlin to the distribution centers in both Los Angeles and Seattle. This figure also gives in brackets the maximum amount that can be shipped through each shipping lane. In the weeks following the crisis described in Section 6.2, the distribution center in Los Angeles has successfully replenished its inventory. Therefore, Karl Schmidt (the supply chain manager for the BMZ Co.) has concluded that it will be sufficient hereafter to ship 130 units per month to Los Angeles and 50 units per month to Seattle. (One unit is a hundred cubic meters of automobile replacement parts.) The Stuttgart factory (node ST in the figure) will allocate 130 units per month and the Berlin factory (node BE) will allocate 50 units per month out of their total production to cover these shipments. However, rather than resuming the past practice of supplying the Los Angeles distribution center from only the Stuttgart factory and supplying the Seattle distribution center from only the Berlin factory, Karl has decided to allow either factory to supply either distribution center. He feels that this additional flexibility is likely to reduce the total shipping cost. The following table gives the shipping cost per unit through each of these shipping lanes. Unit Shipping Cost to Node To From Node ST BE LI BO RO HA NO NY BN LI BO RO HA NO NY BN LA SE $3,200 — — — — — — — — $2,500 — — — — — — — — $2,900 $2,400 — — — — — — — — $2,000 — — — — — — — — — $6,100 $6,800 — — — — — — — — $5,400 $5,900 $6,300 — — — — — — — — $5,700 — — — — — — — — — $3,100 $4,200 $3,400 — — — — — — — $4,000 $3,000 Chapter 6 Problems 233 Karl wants to determine the shipping plan that will minimize the total shipping cost. a. Formulate a network model for this problem as a minimum-cost flow problem by inserting all the necessary data into the distribution network shown in Figure 6.9. (Use the format depicted in Figure 6.3 to display these data.) b. Formulate and solve a spreadsheet model for this problem. c. What is the total shipping cost for this optimal solution? 6.7. Reconsider Problem 6.6. Suppose now that, for administrative convenience, management has decided that all 130 units per month needed at the distribution center in Los Angeles must come from the Stuttgart factory (node ST) and all 50 units per month needed at the distribution center in Seattle must come from the Berlin factory (node BE). For each of these distribution centers, Karl Schmidt wants to determine the shipping plan that will minimize the total shipping cost. a. For the distribution center in Los Angeles, formulate a network model for this problem as a minimum-cost flow problem by inserting all the necessary data into the distribution network shown in Figure 6.6. (Use the format depicted in Figure 6.3 to display these data.) b. Formulate and solve a spreadsheet model for the problem formulated in part a. c. For the distribution center in Seattle, draw its distribution network emanating from the Berlin factory at node BE. d. Repeat part a for the distribution center in Seattle by using the network drawn in part c. e. Formulate and solve a spreadsheet model for the problem formulated in part d. f. Add the total shipping costs obtained in parts b and e. Compare this sum with the total shipping cost obtained in part c of Problem 6.6 (as given in the back of the book). 6.8. Consider the maximum flow problem formulated in Figures 6.7 and 6.8 for the BMZ case study. Redraw Figure 6.7 and insert the optimal shipping quantities (cells D4:D12 in ­Figure 6.8) in parentheses above the respective arcs. Examine the capacities of these arcs. Explain why these arc capacities ensure that the shipping quantities in parentheses must be an optimal solution because the maximum flow cannot exceed 150. 6.9. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 6.3. Briefly describe how a generalization of the model for the maximum flow problem was applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 6.10.* Formulate and solve a spreadsheet model for the maximum flow problem shown next, where node A is the source, node F is the sink, and the arc capacities are the numbers in square brackets shown next to the arcs. B [9] A D [7] [6] [2] [3] F [4] [7] C [9] E [6] 6.11. The diagram depicts a system of aqueducts that originate at three rivers (nodes R1, R2, and R3) and terminate at a major city (node T), where the other nodes are junction points in the system. A D R1 T B E R2 C R3 F Using units of thousands of acre feet, the following tables show the maximum amount of water that can be pumped through each aqueduct per day. To From R1 R2 R3 A B C 75 40 — 65 50 80 — 60 70 To From A B C D E F From\To 60 70 — 45 55 70 — 45 90 D E F T 120 190 130 The city water manager wants to determine a flow plan that will maximize the flow of water to the city. a. Formulate this problem as a maximum flow problem by identifying a source, a sink, and the transshipment nodes, and then drawing the complete network that shows the capacity of each arc. b. Formulate and solve a spreadsheet model for this problem. 234 Chapter Six Network Optimization Problems 6.12. The Texago Corporation has four oil fields, four refineries, and four distribution centers in the locations identified in the next tables. A major strike involving the transportation industries now has sharply curtailed Texago’s capacity to ship oil from the four oil fields to the four refineries and to ship petroleum products from the refineries to the distribution centers. Using units of thousands of barrels of crude oil (and its equivalent in refined products), the following tables show the maximum number of units that can be shipped per day from each oil field to each refinery and from each refinery to each distribution center. Refinery Oil Field Texas California Alaska Middle East New Orleans Charleston Seattle St. Louis 11 5 7 8 7 4 3 9 2 8 12 4 8 7 6 15 Distribution Center Refinery New Orleans Charleston Seattle St. Louis Pittsburgh Atlanta Kansas City San Francisco 5 9 6 4 8 4 12 7 6 11 9 7 9 5 8 7 The Texago management now wants to determine a plan for how many units to ship from each oil field to each refinery and from each refinery to each distribution center that will maximize the total number of units reaching the distribution centers. a. Draw a rough map that shows the location of Texago’s oil fields, refineries, and distribution centers. Add arrows to show the flow of crude oil and then petroleum products through this distribution network. b. Redraw this distribution network by lining up all the nodes representing oil fields in one column, all the nodes representing refineries in a second column, and all the nodes representing distribution centers in a third column. Then add arcs to show the possible flow. c. Use the distribution network from part b to formulate a network model for Texago’s problem as a variant of a maximum flow problem. d. Formulate and solve a spreadsheet model for this problem. 6.13. Read the referenced article that fully describes the management science study summarized in the application vignette presented in Section 6.4. Briefly describe how network optimization models (including for shortest path problems) were applied in this study. Then list the various financial and nonfinancial benefits that resulted from this study. 6.14. Reconsider the Littletown Fire Department problem presented in Section 6.4 and depicted in Figure 6.11. Due to maintenance work on the one-mile road between nodes A and B, a detour currently must be taken that extends the trip between these nodes to four miles. Formulate and solve a spreadsheet model for this revised problem to find the new shortest path from the fire station to the farming community. 6.15. You need to take a trip by car to another town that you have never visited before. Therefore, you are studying a map to determine the shortest route to your destination. Depending on which route you choose, there are five other towns (call them A, B, C, D, E) through which you might pass on the way. The map shows the mileage along each road that directly connects two towns without any intervening towns. These numbers are summarized in the following table, where a dash indicates that there is no road directly connecting these two towns without going through any other towns. Miles between Adjacent Towns Town A B C D E Destination Origin A B C D E 40 60 10 50 — 20 — 70 55 — — — 40 50 10 — — — — 60 80 a. Formulate a network model for this problem as a shortest path problem by drawing a network where nodes represent towns, links represent roads, and numbers indicate the length of each link in miles. b. Formulate and solve a spreadsheet model for this problem. c. Use part b to identify your shortest route. d. If each number in the table represented your cost (in dollars) for driving your car from one town to the next, would the answer in part c now give your minimum-cost route? e. If each number in the table represented your time (in minutes) for driving your car from one town to the next, would the answer in part c now give your minimum-time route? 6.16.* At a small but growing airport, the local airline company is purchasing a new tractor for a tractor-trailer train to bring luggage to and from the airplanes. A new mechanized luggage system will be installed in three years, so the tractor will not be needed after that. However, because it will receive heavy use, so that the running and maintenance costs will increase rapidly as it ages, it may still be more economical to replace the tractor after one or two years. The next table gives the total net discounted cost associated with purchasing a tractor (purchase price minus trade-in allowance, plus running and maintenance costs) at the end of year i and trading it in at the end of year j (where year 0 is now). Case 6-1 Aiding Allies 235 j 3 $8,000 $18,000 10,000 $31,000 21,000 12,000 3.4 SE 3.2 B E 3.6 LN 3.3 2 4. Management wishes to determine at what times (if any) the tractor should be replaced to minimize the total cost for the tractor(s) over three years. a. Formulate a network model for this problem as a shortest path problem. b. Formulate and solve a spreadsheet model for this problem. 6.17. One of Speedy Airlines’s flights is about to take off from Seattle for a nonstop flight to London. There is some flexibility in choosing the precise route to be taken, depending upon weather conditions. The following network depicts the possible routes under consideration, where SE and LN are Seattle and London, respectively, and the other nodes represent various intermediate locations. The winds along each arc greatly affect the flying time (and so the fuel consumption). Based on current meteorological 4.7 3.6 3.8 2 4.6 1 D 3.4 i 0 1 2 3.5 A 3.5 C 3.4 F reports, the flying times (in hours) for this particular flight are shown next to the arcs. Because the fuel consumed is so expensive, the management of Speedy Airlines has established a policy of choosing the route that minimizes the total flight time. a. What plays the role of distances in interpreting this problem to be a shortest path problem? b. Formulate and solve a spreadsheet model for this problem. Case 6-1 Aiding Allies Commander Votachev steps into the cold October night and deeply inhales the smoke from his cigarette, savoring its warmth. He surveys the destruction surrounding him—shattered windows, burning buildings, torn roads—and smiles. His two years of work training revolutionaries east of the Ural Mountains has proven successful; his troops now occupy seven strategically important cities in the Russian Federation: Kazan, Perm, Yekaterinburg, Ufa, Samara, Saratov, and Orenburg. His siege is not yet over, however. He looks to the west. Given the political and economic confusion in the Russian Federation at this time, he knows that his troops will be able to conquer Saint Petersburg and Moscow shortly. Commander Votachev will then be able to rule with the wisdom and control exhibited by his communist predecessors Lenin and Stalin. Across the Pacific Ocean, a meeting of the top security and foreign policy advisors of the United States is in progress at the White House. The president has recently been briefed about the communist revolution masterminded by Commander Votachev and is determining a plan of action. The president reflects upon a similar October long ago in 1917, and he fears the possibility of a new age of radical Communist rule accompanied by chaos, bloodshed, escalating tensions, and possibly nuclear war. He therefore decides that the United States needs to respond and to respond quickly. Moscow has requested assistance from the United States military, and the president plans to send troops and supplies immediately. The president turns to General Lankletter and asks him to describe the preparations being taken in the United States to send the necessary troops and supplies to the Russian Federation. General Lankletter informs the president that along with troops, weapons, ammunition, fuel, and supplies, aircraft, ships, and vehicles are being assembled at two port cities with airfields: Boston and Jacksonville. The aircraft and ships will transfer all troops and cargo across the Atlantic Ocean to the Eurasian continent. The general hands the president a list of the types of aircraft, ships, and vehicles being assembled along with a description of each type. The list is shown next. Transportation Type Name Capacity Speed Aircraft Ship Vehicle C-17 Globemaster Transport Palletized Load System Truck 150 tons 240 tons 16,000 kilograms 400 miles per hour 35 miles per hour 60 miles per hour 236 Chapter Six Network Optimization Problems All aircraft, ships, and vehicles are able to carry both troops and cargo. Once an aircraft or ship arrives in Europe, it stays there to support the armed forces. The president then turns to Tabitha Neal, who has been negotiating with the NATO countries for the last several hours to use their ports and airfields as stops to refuel and resupply before heading to the Russian Federation. She informs the president that the following ports and airfields in the NATO countries will be made available to the U.S. military. Ports Airfields Napoli Hamburg Rotterdam London Berlin Istanbul The president stands and walks to the map of the world projected on a large screen in the middle of the room. He maps the progress of troops and cargo from the United States to three strategic cities in the Russian Federation that have not yet been seized by Commander Votachev. The three cities are Saint Petersburg, Moscow, and Rostov. He explains that the troops and cargo will be used both to defend the Russian cities and to launch a counter attack against Votachev to recapture the cities he currently occupies. (The map is shown at the end of the case.) The president also explains that all Globemasters and transports leave Boston or Jacksonville. All transports that have traveled across the Atlantic must dock at one of the NATO ports to unload. Palletized load system trucks brought over in the transports will then carry all troops and materials unloaded from the ships at the NATO ports to the three strategic Russian cities not yet seized by Votachev. All Globemasters that have traveled across the Atlantic must land at one of the NATO airfields for refueling. The planes will then carry all troops and cargo from the NATO airfields to the three Russian cities. a. Draw a network showing the different routes troops and supplies may take to reach the Russian Federation from the United States. b. Moscow and Washington do not know when Commander Votachev will launch his next attack. Leaders from the two countries therefore have agreed that troops should reach each of the three strategic Russian cities as quickly as possible. The president has determined that the situation is so dire that cost is no object—as many Globemasters, transports, and trucks as are necessary will be used to transfer troops and cargo from the United States to Saint Petersburg, Moscow, and Rostov. Therefore, no limitations exist on the number of troops and amount of cargo that can be transferred between any cities. The president has been given the information in the next table about the length of the available routes between cities. Given the distance and the speed of the transportation used between each pair of cities, how can the president most quickly move troops from the United States to each of the three strategic Russian cities? Highlight the path(s) on the network. How long will it take troops and supplies to reach Saint Petersburg? Moscow? Rostov? From To (Kilometers) Boston Boston Boston Boston Boston Boston Jacksonville Jacksonville Jacksonville Jacksonville Jacksonville Jacksonville Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Moscow Moscow Moscow Moscow Moscow Moscow Rostov Rostov Rostov Rostov Rostov Rostov 7,250 km 8,250 8,300 6,200 6,900 7,950 9,200 9,800 10,100 7,900 8,900 9,400 1,280 1,880 2,040 1,980 2,200 2,970 1,600 2,120 1,700 2,300 2,450 2,890 1,730 2,470 990 2,860 2,760 2,800 c. The president encounters only one problem with his first plan: He has to sell the military deployment to Congress. Under the War Powers Act, the president is required to consult with Congress before introducing troops into hostilities or situations where hostilities will occur. If Congress does not give authorization to the president for such use of troops, the president must withdraw troops after 60 days. Congress also has the power to decrease the 60-day time period by passing a concurrent resolution. The president knows that Congress will not authorize significant spending for another country’s war, especially when voters have paid so much attention to decreasing the national debt. He therefore decides that he needs to find a way to get the needed troops and supplies to Saint Petersburg, Moscow, and Rostov at the minimum cost. Each Russian city has contacted Washington to communicate the number of troops and supplies the city needs at a minimum for reinforcement. After analyzing the requests, General Lankletter has converted the requests from numbers of troops, gallons of gasoline, and so on, to tons of cargo for easier planning. The requirements are listed next. City Saint Petersburg Moscow Rostov Requirements 320,000 tons 440,000 tons 240,000 tons Case 6-1 Aiding Allies 237 Both in Boston and Jacksonville, there are 500,000 tons of the necessary cargo available. When the United States decides to send a plane, ship, or truck between two cities, several costs occur: fuel costs, labor costs, maintenance costs, and appropriate port or airfield taxes and tariffs. These costs are listed next. From To Cost Boston Boston Boston Boston Boston Boston Jacksonville Jacksonville Jacksonville Jacksonville Jacksonville Jacksonville Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Berlin Hamburg Istanbul London Rotterdam Napoli Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Saint Petersburg Moscow Moscow Moscow Moscow Moscow Moscow Rostov Rostov Rostov Rostov Rostov Rostov $50,000 per Globemaster $30,000 per transport $55,000 per Globemaster $45,000 per Globemaster $30,000 per transport $32,000 per transport $57,000 per Globemaster $48,000 per transport $61,000 per Globemaster $49,000 per Globemaster $44,000 per transport $56,000 per transport $24,000 per Globemaster $3,000 per truck $28,000 per Globemaster $22,000 per Globemaster $3,000 per truck $5,000 per truck $22,000 per Globemaster $4,000 per truck $25,000 per Globemaster $19,000 per Globemaster $5,000 per truck $5,000 per truck $23,000 per Globemaster $7,000 per truck $2,000 per Globemaster $4,000 per Globemaster $8,000 per truck $9,000 per truck The president faces a number of restrictions when trying to satisfy the requirements. Early winter weather in northern Russia has brought a deep freeze with much snow. Therefore, General Lankletter is opposed to sending truck convoys in the area. He convinces the president to supply Saint Petersburg only through the air. Moreover, the truck routes into Rostov are quite limited, so that from each port, at most 2,500 trucks can be sent to Rostov. The Ukrainian government is very sensitive about American airplanes flying through its air space. It restricts the U.S. military to at most 200 flights from Berlin to Rostov and to at most 200 flights from London to Rostov. (The U.S. military does not want to fly around the Ukraine and is thus restricted by the Ukrainian limitations.) How does the president satisfy each Russian city’s military requirements at minimum cost? Highlight the path to be used between the United States and the Russian Federation on the network. d. Once the president releases the number of planes, ships, and trucks that will travel between the United States and the Russian Federation, Tabitha Neal contacts each of the American cities and NATO countries to indicate the number of planes to expect at the airfields, the number of ships to expect at the docks, and the number of trucks to expect traveling across the roads. Unfortunately, Tabitha learns that several additional restrictions exist that cannot be immediately eliminated. Because of airfield congestion and unalterable flight schedules, only a limited number of planes may be sent between any two cities. These plane limitations are given below. From To Boston Boston Boston Jacksonville Jacksonville Jacksonville Berlin Istanbul London Berlin Istanbul London Berlin Istanbul London Berlin Istanbul London Berlin Istanbul London Saint Petersburg Saint Petersburg Saint Petersburg Moscow Moscow Moscow Rostov Rostov Rostov Maximum Number of Airplanes 300 500 500 500 700 600 500 0 1,000 300 100 200 0 900 100 In addition, because some countries fear that citizens will become alarmed if too many military trucks travel the public highways, they object to a large number of trucks traveling through their countries. These objections mean that a limited number of trucks are able to travel between certain ports and Russian cities. These limitations are listed below. From To Rotterdam Rotterdam Hamburg Hamburg Napoli Napoli Moscow Rostov Moscow Rostov Moscow Rostov Maximum Number of Trucks 600 750 700 500 1,500 1,400 Tabitha learns that all shipping lanes have no capacity limits due to the American control of the Atlantic Ocean. The president realizes that due to all the restrictions, he will not be able to satisfy all the reinforcement requirements of the three Russian cities. He decides to disregard the cost issue and instead to maximize the total amount of cargo he can get to the Russian cities. How does the president maximize the total amount of cargo that reaches the Russian Federation? Highlight the path(s) used between the United States and the Russian Federation on the network. 238 Chapter Six Network Optimization Problems St. Petersburg Boston London Jacksonville Hamburg Berlin Rotterdam Perm Kazan Yekaterinburg Moscow Ufa Samara Orenburg Saratov Rostov Napoli Istanbul Case 6-2 Money in Motion Jake Nguyen runs a nervous hand through his once finely combed hair. He loosens his once perfectly knotted silk tie. And he rubs his sweaty hands across his once immaculately pressed trousers. Today has certainly not been a good day. Over the past few months, Jake had heard whispers circulating from Wall Street—whispers from the lips of investment bankers and stockbrokers famous for their outspokenness. They had whispered about a coming Japanese economic collapse— whispered because they had believed that publicly vocalizing their fears would hasten the collapse. And, today, their very fears have come true. Jake and his colleagues gather around a small television dedicated exclusively to the Bloomberg channel. Jake stares in disbelief as he listens to the horrors taking place in the Japanese market. And the Japanese market is taking the financial markets in all other East Asian countries with it on its tailspin. He goes numb. As manager of Asian foreign investment for Grant Hill Associates, a small West Coast investment boutique specializing in currency trading, Jake bears personal responsibility for any negative impacts of the collapse. And Grant Hill Associates will experience negative impacts. Jake had not heeded the whispered warnings of a Japanese collapse. Instead, he had greatly increased the stake Grant Hill Associates held in the Japanese market. Because the Japanese market had performed better than expected over the past year, Jake had increased investments in Japan from $2.5 million to $15 million only one month ago. At that time, one dollar was worth 80 yen. No longer. Jake realizes that today’s devaluation of the yen means that one dollar is worth 125 yen. He will be able to liquidate these investments without any loss in yen, but now the dollar loss when converting back into U.S. currency would be huge. He takes a deep breath, closes his eyes, and mentally prepares himself for serious damage control. Jake’s meditation is interrupted by a booming voice calling for him from a large, corner office. Grant Hill, the president of Grant Hill Associates, yells, “Nguyen, get the hell in here!” Jake jumps and looks reluctantly toward the corner office hiding the furious Grant Hill. He smooths his hair, tightens his tie, and walks briskly into the office. Grant Hill meets Jake’s eyes upon his entrance and continues yelling, “I don’t want one word out of you, Nguyen! No excuses; just fix this debacle! Get all of our money out of Japan! My gut tells me this is only the beginning! Get the money into safe U.S. bonds! NOW! And don’t forget to get our cash positions out of Indonesia and Malaysia ASAP with it!” Jake has enough common sense to say nothing. He nods his head, turns on his heels, and practically runs out of the office. Safely back at his desk, Jake begins formulating a plan to move the investments out of Japan, Indonesia, and Malaysia. His experiences investing in foreign markets have taught him that when playing with millions of dollars, how he gets money out of a foreign market is almost as important as when he gets money out of the market. The banking partners of Grant Hill Associates charge different transaction fees for converting one currency into another one and wiring large sums of money around the globe. And now, to make matters worse, the governments in East Asia have imposed very tight limits on the amount of money an individual or a company can exchange from the domestic currency into a particular foreign currency and withdraw it from the country. The goal of this dramatic measure is to reduce the outflow of foreign investments out of those countries to prevent a complete collapse of the economies in the region. Because of Grant Hill Associates’ cash holdings of 10.5 billion Indonesian rupiahs and 28 million Malaysian ringgits, along with the holdings in yen, it is not clear how these holdings should be converted back into dollars. Case 6-2 Money in Motion 239 Jake wants to find the most cost-effective method to convert these holdings into dollars. On his company’s website, he always can find on-the-minute exchange rates for most currencies in the world (see Table 1). The table states that, for example, 1 Japanese yen equals 0.008 U.S. dollars. By making a few phone calls, he discovers the transaction costs (expressed as a percentage of the currency being converted) his company must pay for large currency transactions during these critical times (see Table 2). Jake notes that exchanging one currency for another one results in the same transaction cost as a reverse conversion. Finally, Jake finds out the maximum amounts of domestic currencies (expressed as the current equivalent of thousands of dollars) his company is allowed to convert into other currencies in Japan, Indonesia, and Malaysia (see Table 3). a. Formulate Jake’s problem as a minimum-cost flow problem, and draw the network for his problem. Identify the supply and demand nodes for the network. b. Which currency transactions must Jake perform to convert the investments from yens, rupiahs, and ringgits into U.S. dollars to ensure that Grant Hill Associates has the maximum dollar amount after all transactions have occurred? How much money does Jake have to invest in U.S. bonds? c. The World Trade Organization forbids transaction limits because they promote protectionism. If no transaction limits exist, what method should Jake use to convert the Asian holdings from the respective currencies into dollars? d. In response to the World Trade Organization’s mandate forbidding transaction limits, the Indonesian government introduces a new tax to protect its currency that leads to a 500 percent increase in transaction costs for transactions of rupiahs. Given these new transaction costs but no transaction limits, what currency transactions should Jake perform to convert the Asian holdings from the respective currencies into dollars? e. Jake realizes that his analysis is incomplete because he has not included all aspects that might influence his planned currency exchanges. Describe other factors that Jake should examine before he makes his final decision. TABLE 1 Currency Exchange Rates To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso 1 50 1 0.04 0.0008 1 0.008 0.00016 0.2 1 0.01 0.0002 0.25 1.25 1 0.0064 0.000128 0.16 0.8 0.64 1 0.0048 0.000096 0.12 0.6 0.48 0.75 1 0.0768 0.001536 1.92 9.6 7.68 12 16 1 Japanese yen Indonesian rupiah Malaysian ringgit U.S. dollar Canadian dollar European euro English pound Mexican peso TABLE 2 Transaction Cost (Percent) To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso — 0.5 — 0.5 0.7 — 0.4 0.5 0.7 — 0.4 0.3 0.7 0.05 — 0.4 0.3 0.4 0.1 0.2 — 0.25 0.75 0.45 0.1 0.1 0.05 — 0.5 0.75 0.5 0.1 0.1 0.5 0.5 — Yen Rupiah Ringgit U.S. dollar Canadian dollar Euro Pound Peso TABLE 3 Transaction Limits in Equivalent of 1,000 Dollars To From Yen Rupiah Ringgit U.S. Dollar Canadian Dollar Euro Pound Peso Yen Rupiah Ringgit — 5,000 3,000 5,000 — 4,500 5,000 2,000 — 2,000 200 1,500 2,000 200 1,500 2,000 1,000 2,500 2,000 500 1,000 4,000 200 1,000 240 Chapter Six Network Optimization Problems Case 6-3 Airline Scheduling Rachel Cook is very concerned. Until recently, she has always had the golden touch, having successfully launched two startup companies that made her a very wealthy woman. However, the timing could not have been worse for her latest start-up— a regional airline called Northwest Commuter that operates on the west coast of the United States. All had been well at the beginning. Four airplanes had been leased and the company had become fairly well established as a no-frills airline providing low-cost commuter flights between the west coast cities of Seattle, Portland, and San Francisco. Achieving fast turnaround times between flights had given Northwest Commuter an important competitive advantage. Then the cost of jet fuel began spiraling upward and the company began going heavily into the red (like so many other airlines at the time). Although some of the flights were still profitable, others were losing a lot of money. Fortunately, jet fuel costs now are starting to come down, but it has become clear to Rachel that she needs to find new ways for Northwest Commuter to become a more efficient airline. In particular, she wants to start by dropping unprofitable flights and then identifying the most profitable combination of flights (including some new ones) for the coming year that could feasibly be flown by the four airplanes. A little over a decade ago, Rachel had been an honor graduate of a leading MBA program. She had enjoyed the management science course she took then and she has decided to apply spreadsheet modeling to analyze her problem. The leasing cost for each airplane is $30,000 per day. At the end of the day, an airplane might remain in the city where it landed on its last flight. Another option is to fly empty overnight to another city to be ready to start a flight from there the next morning. The cost of this latter option is $5,000. Flight Number From To Depart Arrive 1257 2576 8312 1109 3752 2498 8787 8423 7922 5623 2448 1842 3487 4361 4299 1288 3335 9348 7400 7328 6386 6923 Seattle Seattle Seattle Seattle Seattle Seattle Seattle Seattle Portland Portland Portland Portland Portland Portland Portland San Francisco San Francisco San Francisco San Francisco San Francisco San Francisco San Francisco San Francisco Portland San Francisco San Francisco San Francisco Portland San Francisco Portland Seattle San Francisco San Francisco Seattle Seattle San Francisco Seattle Seattle Portland Seattle Seattle Portland Portland Seattle 8:00 AM 9:30 AM 9:30 AM 12:00 PM 2:30 PM 3:00 PM 5:00 PM 6:30 PM 9:00 AM 9:30 AM 11:00 AM 12:00 PM 2:00 PM 4:00 PM 6:00 PM 8:00 AM 8:30 AM 10:30 AM 12:00 PM 12:00 PM 4:00 PM 5:00 PM 10:00 AM 10:30 AM 11:30 AM 2:00 PM 4:30 PM 4:00 PM 7:00 PM 7:30 PM 10:00 AM 11:00 AM 12:30 PM 1:00 PM 3:00 PM 5:30 PM 7:00 PM 10:00 AM 10:00 AM 12:30 PM 2:00 PM 1:30 PM 5:30 PM 7:00 PM The accompanying table shows the 22 possible flights that are being considered for the coming year. The last column gives the estimated net revenue (in thousands of dollars) for each flight, given the average number of passengers anticipated for that flight. a. To simplify the analysis, assume for now that there is virtually no turnaround time between flights so the next flight Expected Revenue ($000) 37 20 25 27 23 18 29 27 20 23 19 21 22 29 27 32 26 24 27 24 28 32 can begin as soon as the current flight ends. (If an immediate next flight is not available, the airplane would wait until the next scheduled flight from that city.) Develop a network that displays some of the feasible routings of the flights. (Hint: Include separate nodes for each half hour between 8:00 AM and 7:30 PM in each city.) Then develop and apply the corresponding spreadsheet model that finds the feasible combination of flights that maximizes the total profit. Case 6-4 Broadcasting the Olympic Games 241 b. Rachel is considering leasing additional airplanes to achieve economies of scale. The leasing cost of each one again would be $30,000 per day. Perform what-if analysis to determine whether it would be worthwhile to have 5, 6, or 7 airplanes instead of 4. c. Now repeat part a under the more realistic assumption that there is a minimum turnaround time of 30 minutes on the ground for unloading and loading passengers between the arrival of a flight and the departure of the next flight by the same airplane. (Most airlines use a considerably longer turnaround time.) Does this change the number of flights that can be flown? d. Rachel now is considering having each of the four airplanes carry freight instead of flying empty if it flies overnight to another city. Instead of a cost of $5,000, this would result in net revenue of $5,000. Adapt the spreadsheet model used in part c to find the feasible combination of flights that maximizes the total profit. Does this change the number of airplanes that fly overnight to another city? Case 6-4 Broadcasting the Olympic Games The management of the WBC television network has been celebrating for days. What a coup! After several unsuccessful attempts in recent decades, they finally have hit the big jackpot. They have won the bidding war to gain the rights to broadcast the next Summer Olympic Games! The price was enormous. However, the advertising income also will be huge. Even if the network loses some money in the process, the gain in prestige should make it all worthwhile. After all, the entire world follows these games closely every four years. Now the entire world receiving the feed of the broadcast from the WBC network will learn what a preeminent network it is. However, reality also is setting in for WBC management. Telecasting the entire Olympic Games will be an enormously complex task. Many different sporting events will be occurring simultaneously in far-flung venues. An unprecedented amount of live television and live-on-the-Internet coverage of the various sporting events needs to be planned. Due to the high amount of bandwidth that will be required to transmit the coverage of the games back to its home studios, WBC needs to upgrade its computer network. It operates a private computer network as shown in the network diagram below. The games will be held near node A. WBC’s home studios are located at node G. At peak times, coverage of the games will require 35 GB/s (35 gigabytes of data per second) to be sent through the network from node A to node G. The capacity of each link in the network is shown in the diagram below (in GB/s). WBC can divide the transmission and route it through multiple paths of the network from A to G, so long as the total bandwidth required on each link does not exceed the capacity of that link. 7 B 13 5 A E 9 D C a. By utilizing the entire computer network, what is the maximum bandwidth available (in GB/s) for transmission from the general site of the Olympic Games (node A) to the home studios (node G)? Set up and solve a linear programming spreadsheet model. b. WBC would like to expand the capacity of the network so it can handle the peak requirement of 35 GB/s from the Olympics site (A) to the home studios (G). WBC can increase the capacity of each link of the computer network by installing 9 4 6 G 3 10 6 F 12 8 additional fiber optic cables. The next table shows the existing capacity of each network segment (in GB/s), the maximum additional capacity that can be added (in GB/s), and the cost to increase the capacity (in millions of dollars per unit GB/s added). Make a copy of the spreadsheet model used to solve part a and make any revisions necessary to solve this new problem of handling the peak requirement of 35 GB/s. Note: This case will be continued in the next chapter ( Case 7-4 ), so we suggest that you save your spreadsheet model from part b. 242 Chapter Six Network Optimization Problems Network Segment From To Existing Capacity (GB/s) Maximum Additional Capacity (GB/s) Cost per GB/s of Additional Capacity ($million) A A A B B B C D D E E F B C D D E F D E G F G G 13 6 10 9 5 7 8 3 12 4 6 9 6 4 3 4 5 3 5 2 5 2 4 5 2.8 2.5 2.8 2.5 3.1 1.6 3.9 2.8 1.6 4.6 2.9 1.8 Additional Cases Additional cases for this chapter also are available at the University of Western Ontario Ivey School of Business website, cases. ivey.uwo.ca/cases, in the segment of the CaseMate area designated for this book. Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions Learning Objectives After completing this chapter, you should be able to 1. Describe how binary decision variables are used to represent yes-or-no decisions. 2. Use binary decision variables to formulate constraints for mutually exclusive alternatives and contingent decisions. 3. Formulate a binary integer programming model for the selection of projects. 4. Formulate a binary integer programming model for the selection of sites for facilities. 5. Formulate a binary integer programming model for crew scheduling in the travel industry. 6. Formulate other basic binary integer programming models from a description of the problems. 7. Use mixed binary integer programming to deal with setup costs for initiating the production of a product. The preceding chapters have considered various kinds of problems where decisions need to be made about how much to do of various activities. Thus, the decision variables in the resulting model represent the level of the corresponding activities. We turn now to a common type of problem where, instead of how-much decisions, the decisions to be made are yes-or-no decisions. A yes-or-no decision arises when a particular option is being considered and the only possible choices are yes, go ahead with this option, or no, decline this option. The natural choice of a decision variable for a yes-or-no decision is a binary variable. Binary variables are variables whose only possible values are 0 and 1. Thus, when representing a yes-or-no decision, a binary decision variable is assigned a value of 1 for choosing yes and a value of 0 for choosing no. Models that fit linear programming except that they use binary decision variables are called binary integer programming (BIP) models. (We hereafter will use the BIP abbreviation.) A pure BIP model is one where all the variables are binary variables, whereas a mixed BIP model is one where only some of the variables are binary variables. A BIP model can be considered to be a special type of integer programming model. A general integer programming model is simply a linear programming model except for also having constraints that some or all of the decision variables must have integer values (0, 1, 2, . . .). A BIP model further restricts these integer values to be only 0 or 1. However, BIP problems are quite different from general integer programming problems because of the difference in the nature of the decisions involved. Like linear programming problems, general integer programming problems involve how-much decisions, but where these decisions make sense only if they have integer values. For example, the TBA Airlines problem presented in Section 3.2 is a general integer programming problem because it is 243 244 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions essentially a linear programming problem except that its how-much decisions (how many small airplanes and how many large airplanes to purchase) only make sense if they have integer values. By contrast, pure BIP problems involve yes-or-no decisions instead of how-much decisions. Mixed BIP problems involve both types of decisions. The first four sections of the chapter present a variety of examples of pure BIP problems. Section 7.5 provides an example of a mixed BIP problem. For clarity, all of these examples have been kept quite small, so sometimes they could be solved by simply enumerating and evaluating the options individually without even needing BIP. However, they are illustrative of the sometimes vastly larger problems that often arise in practice and require the use of BIP. The preceding chapters already have focused on problems involving how-much decisions and how such techniques as linear programming or integer programming can be used to a­ nalyze these problems. Therefore, this chapter will be devoted instead to problems involving yes-or-no decisions and how BIP models can be used to analyze this special category of problems. BIP problems arise with considerable frequency in a wide variety of applications. To illustrate this, we begin with a case study and then present some more examples in the subsequent sections. One of the supplements to this chapter at www.mhhe.com/Hillier6e also provides additional formulation examples for BIP problems. You will see throughout this chapter that BIP problems can be formulated on a spreadsheet just as readily as linear programming problems. The Solver also can solve BIP problems of modest size. You normally will have no problem solving the small BIP problems found in this book, but Solver may fail on somewhat larger problems. To provide some perspective on this issue, we include another supplement at www.mhhe.com/Hillier6e that is entitled Some Perspectives on Solving Binary Integer Programming Problems. The algorithms available for solving BIP problems (including the one used by Solver) are not nearly as efficient as those for linear programming, so this supplement discusses some of the difficulties and pitfalls involved in solving large BIP problems. One option with any large problem that fits linear programming except that it has decision variables that are restricted to integer values (but not necessarily just 0 and 1) is to ignore the integer constraints and then to round the solution obtained to integer values. This is a reasonable option in some cases but not in others. The supplement emphasizes that this is a particularly dangerous shortcut with BIP problems. 7.1 A CASE STUDY: THE CALIFORNIA MANUFACTURING CO. PROBLEM The top management of the California Manufacturing Company wants to develop a plan for the expansion of the company. Therefore, a management science study will be conducted to help guide the decisions that need to be made. The president of the company, Armando Ortega, is about to meet with the company’s top management scientist, Steve Chan, to discuss the study that management wants done. Let’s eavesdrop on this meeting. Armando Ortega (president): OK, Steve, here is the situation. With our growing business, we are strongly considering building a new factory. Maybe even two. The factory needs to be close to a large, skilled labor force, so we are looking at Los Angeles and San Francisco as the potential sites. We also are considering building one new warehouse. Not more than one. This warehouse would make sense in saving shipping costs only if it is in the same city as a new factory. Either Los Angeles or San Francisco. If we decide not to build a new factory at all, we definitely don’t want the warehouse either. Is this clear, so far? Steve Chan (management scientist): Yes, Armando, I understand, What are your criteria for making these decisions? Armando Ortega: Well, all the other members of top management have joined me in addressing this issue. We have concluded that these two potential sites are very comparable on nonfinancial grounds. Therefore, we feel that these decisions should be based mainly on financial considerations. We have $10 million of capital available for this expansion and we want it to go as far as possible in improving our bottom line. Which feasible combination of investments in factories and warehouses in which locations will be most profitable for the company in the long run? In your language, we want to maximize the total net present value of these investments. An Application Vignette With headquarters in Houston, Texas, Waste Management, Inc. (a Fortune 100 company), is the leading provider of comprehensive waste-management services in North America. Its network of operations includes hundreds of active landfill disposal sites, recycling plants, transfer stations, and collection operations (depots) to provide services to over 25 million residential commercial customers throughout the United States and Canada. The company’s collection-and-transfer vehicles need to follow tens of thousands of daily routes. With an annual operating cost of approximately $120,000 per vehicle, management wanted to have a comprehensive route-management system that would make every route as profitable and efficient as possible. Therefore, a management science team that included a number of consultants was formed to attack this problem. The heart of the route-management system developed by this team is a huge mixed BIP model that optimizes the routes assigned to the respective collection-and-transfer vehicles. Although the What is the most profitable combination of investments? objective function takes several factors into account, the primary goal is the minimization of total travel time. The main decision variables are binary variables that equal 1 if the route assigned to a particular vehicle includes a particular possible leg and that equal 0 otherwise. A geographical information system (GIS) provides the data about the distance and time required to go between any two points. All of this is imbedded within a Web-based Java application that is integrated with the company’s other systems. It is estimated that the implementation of this comprehensive route-management system increased the company’s cash flow by $648 million over a five-year period, largely because of savings of $498 million in operational expenses over this same period. It also is providing better customer service. Source: S. Sahoo, S. Kim, B.-I. Kim, B. Krass, and A. Popov, Jr., “Routing Optimization for Waste Management,” Interfaces 35, no. 1 (January–February 2005), pp. 24–36. (A link to this article is provided at www.mhhe.com/Hillier6e.) Steve Chan: That’s very clear. It sounds like a classical management science problem. Armando Ortega: That’s why I called you in, Steve. I would like you to conduct a quick management science study to determine the most profitable combination of investments. I also would like you to take a look at the amount of capital being made available and its effect on how much profit we can get from these investments. The decision to make $10 million available is only a tentative one. That amount is stretching us, because we now are investigating some other interesting project proposals that would require quite a bit of capital, so we would prefer to use less than $10 million on these particular investments if the last few million don’t buy us much. On the other hand, this expansion into either Los Angeles or San Francisco, or maybe both of these key cities, is our number one priority. It will have a real positive impact on the future of this company. So we are willing to go out and raise some more capital if it would give us a lot of bang for the buck. Therefore, we also would like you to do some what-if analysis to tell us what the effect would be if we were to change the amount of capital being made available to anything between $5 million and $15 million. Steve Chan: Sure, Armando, we do that kind of what-if analysis all the time. We refer to it as sensitivity analysis because it involves checking how sensitive the outcome is to the amount of capital being made available. Armando Ortega: Good. Now, Steve, I need your input within the next couple weeks. Can you do it? Steve Chan: Well, Armando, as usual, the one question is whether we can gather all the necessary data that quickly. We’ll need to get good estimates of the net present value of each of the possible investments. I’ll need a lot of help in digging out that information. Armando Ortega: I thought you would say that. I already have my staff working hard on developing those estimates. I can get you together with them this afternoon. Steve Chan: Great. I’ll get right on it. Background The California Manufacturing Company is a diversified company with several factories and warehouses throughout California, but none yet in Los Angeles or San Francisco. Because the company is enjoying increasing sales and earnings, management feels that the time may be ripe to expand into one or both of those prime locations. A basic issue is whether to build a new factory in either Los Angeles or San Francisco, or perhaps even in both cities. Management also is considering building at most one new warehouse, but will restrict the choice of location to a city where a new factory is being built. 245 246 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions TABLE 7.1 Data for the California Manufacturing Co. Problem Decision Number Yes-or-No Question Decision Variable Net Present Value (Millions) 1 2 3 4 Build a factory in Los Angeles? Build a factory in San Francisco? Build a warehouse in Los Angeles? Build a warehouse in San Francisco? x1 x2 x3 x4 $8 5 6 4 Capital Required (Millions) $6 3 5 2 Capital available: $10 million The decisions to be made are listed in the second column of Table 7.1 in the form of yesor-no questions. In each case, giving an answer of yes to the question corresponds to the decision to make the investment to build the indicated facility (a factory or a warehouse) in the indicated location (Los Angeles or San Francisco). The capital required for the investment is given in the rightmost column, where management has made the tentative decision that the total amount of capital being made available for all the investments is $10 million. (Note that this amount is inadequate for some of the combinations of investments.) The fourth column shows the estimated net present value (net long-run profit considering the time value of money) if the corresponding investment is made. (The net present value is 0 if the investment is not made.) Much of the work of Steve Chan’s management science study (with substantial help from the president’s staff) goes into developing these estimates of the net present values. As specified by the company’s president, Armando Ortega, the objective now is to find the feasible combination of investments that maximizes the total net present value. Introducing Binary Decision Variables for the Yes-or-No Decisions As summarized in the second column of Table 7.1, the problem facing management is to make four interrelated yes-or-no decisions. To formulate a mathematical model for this problem, Steve Chan needs to introduce a decision variable for each of these decisions. Since each decision has just two alternatives, choose yes or choose no, the corresponding decision variable only needs to have two values (one for each alternative). Therefore, Steve uses a binary variable, whose only possible values are 0 and 1, where 1 corresponds to the decision to choose yes and 0 corresponds to choosing no. These decision variables are shown in the second column of Table 7.2. The final two columns give the interpretation of a value of 1 and 0, respectively. Dealing with Interrelationships between the Decisions Recall that management wants no more than one new warehouse to be built. In terms of the corresponding decision variables, x3 and x4, this means that no more than one of these variables is allowed to have the value 1. Therefore, these variables must satisfy the constraint x 3 + x 4 ≤ 1 as part of the mathematical model for the problem. These two alternatives (build a warehouse in Los Angeles or build a warehouse in San Francisco) are referred to as mutually exclusive alternatives because choosing one of these alternatives excludes choosing the other. Groups of two or more mutually exclusive TABLE 7.2 Binary Decision Variables for the California Manufacturing Co. Problem Decision Number Decision Variable Possible Value Interpretation of a Value of 1 Interpretation of a Value of 0 1 2 3 4 x1 x2 x3 x4 0 or 1 0 or 1 0 or 1 0 or 1 Build a factory in Los Angeles Build a factory in San Francisco Build a warehouse in Los Angeles Build a warehouse in San Francisco Do not build this factory Do not build this factory Do not build this warehouse Do not build this warehouse 7.1 A Case Study: The California Manufacturing Co. Problem 247 With a group of mutually exclusive alternatives, only one of the corresponding binary decision variables can equal 1. alternatives arise commonly in BIP problems. For each such group where at most one of the alternatives can be chosen, the constraint on the corresponding binary decision variables has the form shown above, namely, the sum of these variables must be less than or equal to 1. For some groups of mutually exclusive alternatives, management will exclude the possibility of choosing none of the alternatives, in which case the constraint will set the sum of the corresponding binary decision variables equal to 1. The California Manufacturing Co. problem also has another important kind of restriction. Management will allow a warehouse to be built in a particular city only if a factory also is being built in that city. For example, consider the situation for Los Angeles (LA). If decide no, do not build a factory in LA (i.e., if choose x1 = 0), then cannot build a warehouse in LA (i.e., must choose x3 = 0). If decide yes, do build a factory in LA (i.e., if choose x1 = 1), then can either build a warehouse in LA or not (i.e., can choose either x3 = 1 or 0). How can these interrelationships between the factory and warehouse decisions for LA be expressed in a constraint for a mathematical model? The key is to note that, for either value of x1, the permissible value or values of x3 are less than or equal to x1. Since x1 and x3 are binary variables, the constraint x 3 ≤ x 1 forces x3 to take on a permissible value given the value of x1. Exactly the same reasoning leads to x 4 ≤ x 2 One yes-or-no decision is contingent on another yesor-no decision if the first one is allowed to be yes only if the other one is yes. as the corresponding constraint for San Francisco. Just as for Los Angeles, this constraint forces having no warehouse in San Francisco (x4 = 0) if a factory will not be built there (x2 = 0), whereas going ahead with the factory there (x2 = 1) leaves open the decision to build the warehouse there (x4 = 0 or 1). For either city, the warehouse decision is referred to as a contingent decision, because the decision depends on a prior decision regarding whether to build a factory there. In general, one yes-or-no decision is said to be contingent on another yes-or-no decision if it is allowed to be yes only if the other is yes. As above, the mathematical constraint expressing this relationship requires that the binary variable for the former decision must be less than or equal to the binary variable for the latter decision. The rightmost column of Table 7.1 reveals one more interrelationship between the four decisions, namely, that the amount of capital expended on the four facilities under consideration cannot exceed the amount available ($10 million). Therefore, the model needs to include a constraint that requires Capital expended ≤ $10 million Excel Tip: Beware that rounding errors can occur with Excel. Therefore, even when you add a constraint that a changing cell has to be binary, Excel occasionally will return a noninteger value very close to an integer (e.g., 1.23E-10, meaning 0.000000000123). When this happens, you can replace the noninteger value with the proper integer value. How can the amount of capital expended be expressed in terms of the four binary decision variables? To start this process, consider the first yes-or-no decision (build a factory in Los Angeles?). Combining the information in the rightmost column of Table 7.1 and the first row of Table 7.2, $6 million if x 1 = 1 Capital expended on factory in Los Angeles = { if x 1 = 0 0 = $6 million times x 1 By the same reasoning, the amount of capital expended on the other three investment opportunities (in units of millions of dollars) is 3x2, 5x3, and 2x4, respectively. Consequently, Capital expended = 6x 1 + 3x 2 + 5x 3 + 2x 4 (in millions of dollars) Therefore, the constraint becomes 6x 1 + 3x 2 + 5x 3 + 2x 4 ≤ 10 248 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions Excel Tip: In the Solver Options, the Integer Optimality (%) setting (1 ­percent by default) causes Solver to stop solving an integer programming problem when it finds a feasible solution whose objective function value is within the specified percentage of being optimal. (In ­Analytic Solver, this option is called Integer Tolerance and is found in the Engine tab of the Model pane.) This is useful for very large BIP problems since it may enable finding a nearoptimal solution when finding an optimal solution in a reasonable period of time is not possible. For smaller problems (e.g., all the problems in this book), this option should be set to 0 to guarantee finding an optimal solution. Note how helpful the range names are for interpreting this BIP spreadsheet model. The BIP Model As indicated by Armando Ortega in his conversation with Steve Chan, management’s objective is to find the feasible combination of investments that maximizes the total net present value of these investments. Thus, the value of the objective function should be NPV = Total net present value If the investment is made to build a particular facility (so that the corresponding decision variable has a value of 1), the estimated net present value from that investment is given in the fourth column of Table 7.1. If the investment is not made (so the decision variable equals 0), the net present value is 0. Therefore, continuing to use units of millions of dollars, NPV = 8x 1 + 5x 2 + 6x 3 + 4x 4 is the quantity to enter into the objective cell to be maximized. Incorporating the constraints developed in the preceding subsection, the complete BIP model then is shown in Figure 7.1. The format is basically the same as for linear programming models. The one key difference arises when using Solver. Each of the decision variables (cells C18:D18 and C16:D16) is constrained to be binary. In Excel’s Solver, this is accomplished in the Add Constraint dialog box by choosing each range of changing cells as the left-hand side and then choosing bin from the pop-up menu. In Analytic Solver, this is accomplished by selecting each range of changing cells, and then under the Constraints menu on the Analytic Solver ribbon, choose Binary in the Variable Type>Bound submenu. The other constraints shown in Solver (see the lower left-hand side of Figure 7.1) have been made quite intuitive by using the suggestive range names given in the lower right-hand side of the figure. For convenience, the equations entered into the output cells in E12 and D20 use a SUMPRODUCT function that includes C17:D17 and either C11:D11 or C5:D5 because the blanks or ≤ signs in these rows are interpreted as zeroes by Solver. Solver gives the optimal solution shown in C18:D18 and C16:D16 of the spreadsheet, namely, build factories in both Los Angeles and San Francisco, but do not build any warehouses. The objective cell (D20) indicates that the total net present value from building these two factories is estimated to be $13 million. Performing What-If Analysis Solver’s sensitivity report is not available for integer programming problems. Trial-and-error and/or a parameter analysis report can be used to perform what-if analysis for integer programming problems. Now that Steve Chan has used the BIP model to determine what should be done when the amount of capital being made available to these investments is $10 million, his next task is to perform what-if analysis on this amount. Recall that Armando Ortega wants him to determine what the effect would be if this amount were changed to anything else between $5 million and $15 million. In Chapter 5, we described three different methods of performing what-if analysis on a linear programming spreadsheet model when there is a change in a constraint: using trial and error with the spreadsheet, generating a parameter analysis report, or referring to Solver’s sensitivity report. The first two of these can be used on integer programming problems in exactly the same way as for linear programming problems. The third method, however, does not work. The sensitivity report is not available for integer programming problems. This is because the concept of a shadow price and allowable range no longer applies. In contrast to linear programming, the objective function values for an integer programming problem do not change in a predictable manner when the right-hand side of a constraint is changed. It is straightforward to determine the impact of changing the amount of available capital by trial and error. Simply try different values in the data cell CapitalAvailable (G12) and re-solve with Solver. However, a more systematic way to perform this analysis is to generate a parameter analysis report using Analytic Solver. The parameter analysis report works for integer programming models in exactly the same way as it does for linear programming models (as described in Section 5.3 in the subsection entitled Using a Parameter Analysis Report Generated by Analytic Solver to Do Sensitivity Analysis Systematically). 7.1 A Case Study: The California Manufacturing Co. Problem 249 FIGURE 7.1 A spreadsheet formulation of the BIP model for the California Manufacturing Co. case study where the changing cells BuildFactory? (C18:D18) and BuildWarehouse? (C16:D16) give the optimal solution obtained by Solver. A 1 2 B C D E F G California Manufacturing Co. Facility Location Problem 3 NPV ($millions) LA SF 4 Warehouse 6 4 Factory 8 5 5 6 7 8 Capital Required 9 ($millions) LA SF 10 Warehouse 5 2 Factory 6 3 Total Maximum 15 Build? LA SF Warehouses Warehouses 16 Warehouse 0 0 0 <= <= 1 1 11 12 Capital Capital Spent Available <= 9 10 13 14 17 18 Factory <= 1 19 20 Total NPV ($millions) Solver Parameters Set Objective Cell: TotalNPV To: Max By Changing Variable Cells: BuildWarehouse?, BuildFactory? Subject to the Constraints: BuildFactory? = binary BuildWarehouse? = binary BuildWarehouse? <= BuildFactory? CapitalSpent <= CapitalAvailable TotalWarehouses <= MaxWarehouses Solver Options: Make Variables Nonnegative Solving Method: Simplex LP $13 E 10 Capital 11 Spent 12 = SUMPRODUCT(CapitalRequired,Build?) 13 14 Total 15 Warehouses = SUM(BuildWarehouse?) 16 C 20 Range Name Cells Build? BuildWarehouse? BuildFactory? CapitalAvailable CapitalRequired CapitalSpent MaxWarehouses NPV TotalNPV TotalWarehouses C16:D18 C16:D16 C18:D18 G12 C10:D12 E12 G16 C4:D6 D20 E16 D Total NPV ($millions) =SUMPRODUCT(NPV,Build?) After defining CapitalAvailable (G12) as a parameter cell with values ranging from 5 to 15 ($millions), the parameter analysis report shown in Figure 7.2 was generated by executing the series of steps outlined in Section 5.3. Note how Figure 7.2 shows the effect on the optimal solution and the resulting total net present value of varying the amount of capital being made available. 250 Chapter Seven Using Binary Integer Programming to Deal with Yes-or-No Decisions FIGURE 7.2 The parameter analysis report generated by Analytic Solver that shows the effect on the optimal solution and the resulting total net present value of systematically varying the amount of capital being made available for these investments. A 1 CapitalAvailable 2 5 3 6 7 4 8 5 9 6 10 7 11 8 12 9 13 10 14 11 12 15 B BuildWarehouseLA? 0 0 0 0 0 0 0 0 0 1 1 C BuildWarehouseSF? 1 1 1 1 0 0 1 1 1 0 0 D BuildFactoryLA? 0 0 0 0 1 1 1 1 1 1 1 E BuildFactorySF? 1 1 1 1 1 1 1 1 1 1 1 F TotalINPV 9 9 9 9 13 13 17 17 17 19 19 What-if analysis also could be performed on any of the other data cells—NPV (C4:D6), CapitalRequired (C10:D12), and MaxWarehouses (G16)—in a similar way with a parameter analysis report (or by using trial and error with the spreadsheet). However, a careful job was done in developing good estimates of the net present value of each of the possible investments, and there is little uncertainty in the values entered in the other data cells, so Steve Chan decides that further what-if analysis is not needed. Management’s Conclusion Steve Chan’s report is delivered to Armando Ortega within the two-week deadline. The report recommends the plan presented in Figure 7.1 (build a factory in both Los Angeles and San Francisco but no warehouses) if management decides to stick with its tentative decision to make $10 million of capital available for these investments. One advantage of this plan is that it only uses $9 million of this capital, which frees up $1 million of capital for other project proposals currently being investigated. The report also highlights the results shown in F ­ igure 7.2 while emphasizing two points. One is that a heavy penalty would be paid (a reduction in the total net present value from $13 million to $9 million) if the amount of capital being made available were to be reduced below $9 million. The other is that increasing the amount of capital being made available by just $1 million (from $10 million to $11 million) would enable a substantial increase of $4 million in the total net present value (from $13 million to $17 million). However, a much larger further increase in the amount of capital being made available (from $11 million to $14 million) would be needed to enable a considerably smaller further increase in the total net present value (from $17 million to $19 million). Armando Ortega deliberates with other members of top management before making a decision. It is quickly concluded that increasing the amount of capital being made available all the way up to $14 million would be stretching the company’s financial resources too dangerously to justify the relatively small payoff. However, there is considerable discussion of the pros and cons of the two options of using either $9 million or $11 million of capital. Because of the large payoff from the latter option (an additional $4 million in total net present value), management finally decides to adopt the plan presented in row 8 of Figure 7.2. Thus, the company will build new factories in both Los Angeles and San Francisco as well as a new warehouse in San Francisco, with an estimated total net present value of $17 million. However, because of the large capital requirements of this plan, management also decides to defer building the warehouse until the two factories are completed so that their profits can help finance the construction of the warehouse. Review Questions 1. What are the four interrelated decisions that need to be made by the management of the ­California Manufacturing Co.? 2. Why are binary decision variables appropriate to represent these decisions? 3. What is the objective specified by management for this problem? 4. What are the mutually exclusive alternatives in this problem? What is the form of the resulting constraint in the BIP model? 5. What are the contingent decisions in this problem? For each one, what is the form of the resulting constraint in the BIP model? 6. What is the tentative managerial decision on which what-if analysis needs to be performed? 7.2 Using BIP for Project Selection: The Tazer Corp. Problem 251 7.2 USING BIP FOR PROJECT SELECTION: THE TAZER CORP. PROBLEM The California Manufacturing Co. case study focused on four proposed projects: (1) build a factory in Los Angeles, (2) build a factory in San Francisco, (3) build a warehouse in Los Angeles, and (4) build a warehouse in San Francisco. Management needed to make yes-or-no decisions about which of these projects to select. This is typical of many applications of BIP. However, the nature of the projects may vary considerably from one application to the next. Instead of the proposed construction projects in the case study, our next example involves the selection of research and development projects. This example is adapted from both Case 3-7 and its continuation in Section 13.8, but all the relevant information is repeated below. The Tazer Corp. Problem Tazer Corp., a pharmaceutical manufacturing company, is beginning the search for a new breakthrough drug. The following five potential research and development projects have been identified for attempting to develop such a drug. Project Up:Develop a more effective antidepressant that does not cause serious mood swings. Project Stable: Develop a drug that addresses manic depression. Project Choice: Develop a less intrusive birth control method for women. Project Hope: Develop a vaccine to prevent HIV infection. Project Release: Develop a more effective drug to lower blood pressure. The objective is to choose the projects that will maximize the expected profit while satisfying the budget constraint. In contrast to Case 3-7, Tazer management now has concluded that the company cannot devote enough money to research and development to undertake all of these projects. Only $1.2 billion is available, which will be enough for only two or three of the projects. The first row of Table 7.3 shows the amount needed (in millions of dollars) for each of these projects, where the footnote points out the budget constraint that only $1.2 billion is available. The second row estimates each project’s probability of being successful. If a project is successful, it is estimated that the resulting drug would generate the revenue shown in the third row. Thus, the expected revenue (in the statistical sense) from a potential drug is the product of its numbers in the second and third rows, whereas its expected profit is this expected revenue minus the investment given in the first row. These expected profits are shown in the bottom row of Table 7.3. Tazer management now wants to determine which of these projects should be undertaken to maximize their expected total profit. Formulation with Binary Variables Because the decision for each of the five proposed research and development projects is a yesor-no decision, the corresponding decision variables are binary variables. Thus, the decision variable for each project has the following interpretation. 1, if approve the project Decision Variable = {0, if reject the project Let x1, x2, x3, x4, and x5 denote the decision variables for the respective projects in the order in which they are listed in Table 7.3. TABLE 7.3 Data for the Tazer Project Selection Problem Project R&D investment ($million)* Success rate Revenue if successful ($million) Expected profit ($million) 1 (Up) 2 (Stable) 3 (Choice) 4 (Hope) 5 (Release) 400 50% 1,400 300 300 35% 1,200 120 600 35% 2,200 170 500 20% 3,000 100 200 45% 600 70 * Only $1.2 billion is available for R&D investment. An Application Vignette The Midcontinent Independent System Operator, Inc. (MISO) is a nonprofit organization formed in 1998 (and renamed in 2013) to administer the generation and transmission of electricity throughout the midwestern United States. It serves approximately 40 million customers (both individuals and businesses) through its control of approximately 60,000 miles of highvoltage transmission lines and more than 1,000 power plants capable of generating massive amounts of of electricity. This infrastructure spans many midwestern and southern states plus the Canadian province of Manitoba. The key mission of any regional transmission organization is to reliably and efficiently provide the electricity needed by its customers. MISO transformed the way this was done by using mixed binary integer programming to minimize the total cost of providing the needed electricity. Each main binary variable in the model represents a yes-or-no decision about whether a particular power plant should be on during a particular time period. After solving this model, the results are then fed into a linear programming model to set electricity output levels and establish prices for electricity trades. The mixed BIP model is a massive one which began with about 3,300,000 continuous variables, 450,000 binary variables, and 3,900,000 functional constraints. A special technique (Lagrangian relaxation) is used to solve such a huge model. This innovative application of management science yielded savings of approximately $2.5 billion over the four years from 2007 to 2010, with an additional savings of about $7 billion expected through 2020. These dramatic results led to MISO winning the prestigious first prize in the 2011 international competition for the Franz Edelman Award for Achievement in Operations Research and the Management Sciences. Source: B. Carlson and 12 co-authors, “MISO Unlocks Billions in Savings through the Application of Operations Research for Energy and Ancillary Services Markets,” Interfaces 42, no. 1 (January–February 2012), pp. 58–73. (A link to this article is provided at www.mhhe.com/Hillier6e.) If a project is rejected, there is neither any profit nor any loss, whereas the expected profit if a project is approved is given in the bottom row of Table 7.3. Therefore, when using units of millions of dollars, the expected total profit is P = 300x 1 + 120x 2 + 170x 3 + 100x 4 + 70x 5 The objective is to select the projects to approve that will maximize this expected total profit while satisfying the budget constraint. Other than requiring the decision variables to be binary, the budget constraint limiting the total investment to no more than $1.2 billion is the only constraint that has been imposed by Tazer management on the selection of these research and development projects. Referring to the first row of Table 7.3, this constraint can be expressed in terms of the decision variables as 400x 1 + 300x 2 + 600x 3 + 500x 4 + 200x 5 ≤ 1,200 With this background, the stage now is set for formulating a BIP spreadsheet model for this problem. A BIP Spreadsheet Model for the Tazer Problem Figure 7.3 shows a BIP spreadsheet model for this problem. The data in Table 7.3 have been transferred into cells C5:G8. The changing cells are DoProject? (C10:G10) and the objective cell is TotalExpectedProfit (H8). The one functional constraint is depicted in cells H5:J5. In addition, the changing cells DoProject? are constrained to be binary, as shown in the Solver Parameters box. The changing cells DoProject? (C10:G10) in Figure 7.3 show the optimal solution that has been obtained by Solver, namely, Choose Project Up, Project Choice, and Project Release. The objective cell indicates that the resulting total expected profit is $540 million. Review Questions 252 1. How are binary variables used to represent managerial decisions on which projects from a group of proposed projects should be selected for approval? 2. What types of projects are under consideration in the Tazer Corp. problem? 3. What is the objective for this problem? 7.3 Using BIP for The Selection of Sites for Emergency Services Facilities: The Caliente City Problem 253 FIGURE 7.3 A spreadsheet formulation of the BIP model for the Tazer Corp. project selection problem where the changing cells DoProject? (C10:G10) give the optimal solution obtained by Solver. A 1 2 3 4 5 6 7 8 9 10 8 B C D E F G H I J Stable 300 35% 1,200 120 Choice 600 35% 2,200 170 Hope 500 20% 3,000 100 Release 200 45% 600 70 Total 1,200 <= Budget 1,200 0 1 0 1 Tazer Corp. Project Selection Problem R&D Investment ($million) Success Rate Revenue If Successful ($million) Expected Profit ($million) Do Project? B Expected Profit ($million) Up 400 50% 1,400 300 1 C =C7*C6-C5 Range Name Cells Budget J5 DoProject? C10:G10 ExpectedProfit C8:G8 RandDInvestment Revenue C5:G5 C7:G7 4 SuccessRate C6:G6 6 TotalExpectedProfit TotalRandD H8 H5 7 D =D7*D6-D5 E =E7*E6-E5 540 F =F7*F6-F5 G =G7*G6-G5 H 5 8 Total =SUMPRODUCT(RandDInvestment,DoProject?) =SUMPRODUCT(ExpectedProfit,DoProject?) Solver Parameters Set Objective Cell: TotalExpectedProfit To: Max By Changing Variable Cells: DoProject? Subject to the Constraints: DoProject? = binary TotalRandD <= Budget Solver Options: Make Variables Nonnegative Solving Method: Simplex LP 7.3 USING BIP FOR THE SELECTION OF SITES FOR EMERGENCY SERVICES FACILITIES: THE CALIENTE CITY PROBLEM Although the problem encountered in the California Manufacturing Co. case study can be described as a project selection problem (as was done at the beginning of the preceding section), it could just as well have been called a site selection problem. Recall that the company’s management needed to select a site (Los Angeles or San Francisco) for its new factory as well as for its possible new warehouse. For either of the possible sites for the new factory (or the warehouse), there is a yes-or-no decision for whether that site should be